A different method of evaluating $ intfrac1sqrtsin^3(x)cdotsin(x+alpha),dx$
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I was practicing indefinite integrals and came to this integral:
$$
intfrac 1sqrtsin^3(x)cdotsin(x+alpha),dx.
$$
Everywhere there is only one method given for solving this question
which involves dividing numerator and denominator by $sin^2(x).$
However, I wanted to know if there was any other method for solving this or not?
integration trigonometry indefinite-integrals trigonometric-integrals
edited Sep 9 at 14:03
paulplusx
1,059318
asked Sep 4 at 13:29
JIM
385
add a comment |Â
up vote
-3
down vote
favorite
I was practicing indefinite integrals and came to this integral:
$$
intfrac 1sqrtsin^3(x)cdotsin(x+alpha),dx.
$$
Everywhere there is only one method given for solving this question
which involves dividing numerator and denominator by $sin^2(x).$
However, I wanted to know if there was any other method for solving this or not?
integration trigonometry indefinite-integrals trigonometric-integrals
edited Sep 9 at 14:03
paulplusx
1,059318
asked Sep 4 at 13:29
JIM
385
Divide numerator & denominator by $cos^2x$ or $sin^2x$
– lab bhattacharjee
Sep 4 at 13:30
@labbhattacharjee I wanted to know if there was any other method for solving this or not ??
– JIM
Sep 4 at 13:42
add a comment |Â
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
I was practicing indefinite integrals and came to this integral:
$$
intfrac 1sqrtsin^3(x)cdotsin(x+alpha),dx.
$$
Everywhere there is only one method given for solving this question
which involves dividing numerator and denominator by $sin^2(x).$
However, I wanted to know if there was any other method for solving this or not?
integration trigonometry indefinite-integrals trigonometric-integrals
edited Sep 9 at 14:03
paulplusx
1,059318
asked Sep 4 at 13:29
JIM
385
I was practicing indefinite integrals and came to this integral:
$$
intfrac 1sqrtsin^3(x)cdotsin(x+alpha),dx.
$$
Everywhere there is only one method given for solving this question
which involves dividing numerator and denominator by $sin^2(x).$
However, I wanted to know if there was any other method for solving this or not?
integration trigonometry indefinite-integrals trigonometric-integrals
integration trigonometry indefinite-integrals trigonometric-integrals
edited Sep 9 at 14:03
paulplusx
1,059318
asked Sep 4 at 13:29
JIM
385
edited Sep 9 at 14:03
paulplusx
1,059318
asked Sep 4 at 13:29
JIM
385
edited Sep 9 at 14:03
paulplusx
1,059318
edited Sep 9 at 14:03
paulplusx
1,059318
edited Sep 9 at 14:03
paulplusx
1,059318
1,059318
asked Sep 4 at 13:29
JIM
385
asked Sep 4 at 13:29
JIM
385
asked Sep 4 at 13:29
JIM
385
385
Divide numerator & denominator by $cos^2x$ or $sin^2x$
– lab bhattacharjee
Sep 4 at 13:30
@labbhattacharjee I wanted to know if there was any other method for solving this or not ??
– JIM
Sep 4 at 13:42
add a comment |Â
Divide numerator & denominator by $cos^2x$ or $sin^2x$
– lab bhattacharjee
Sep 4 at 13:30
@labbhattacharjee I wanted to know if there was any other method for solving this or not ??
– JIM
Sep 4 at 13:42
Divide numerator & denominator by $cos^2x$ or $sin^2x$
– lab bhattacharjee
Sep 4 at 13:30
Divide numerator & denominator by $cos^2x$ or $sin^2x$
– lab bhattacharjee
Sep 4 at 13:30
@labbhattacharjee I wanted to know if there was any other method for solving this or not ??
– JIM
Sep 4 at 13:42
@labbhattacharjee I wanted to know if there was any other method for solving this or not ??
– JIM
Sep 4 at 13:42
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
So you want to find another method here is one
$$I=displaystyleintdfrac1sin^frac32left(xright)sqrtsinleft(x+aright),mathrmdx$$
$$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$
$$sinleft(xright)=dfrac1cscleft(xright)$$
$$cosleft(xright)=dfraccotleft(xright)cscleft(xright)$$
$$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
$$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
Substitute $u=cotleft(xright)$ Thus $mathrmdx=-dfrac1csc^2left(xright),mathrmdu$
$$I=-displaystyleintdfrac1sqrtsinleft(aright)u+cosleft(aright),mathrmdu$$
$$v=sinleft(aright)u+cosleft(aright)$$
$$mathrmdu=dfrac1sinleft(aright),mathrmdv$$
$$I=-classsteps-nodecssIdsteps-node-3dfrac1sinleft(aright)displaystyleintdfrac1sqrtv,mathrmdv$$
$$I=-dfrac2sqrtvsinleft(aright)$$
$$I=-dfrac2sqrtsinleft(aright)u+cosleft(aright)sinleft(aright)$$
$$u=cotleft(xright)$$
$$I=-dfrac2sqrtsinleft(aright)cotleft(xright)+cosleft(aright)sinleft(aright)$$
$$I=-dfrac2sqrtsinleft(aright)cotleft(xright)+cosleft(aright)sinleft(aright)+C$$
answered Sep 4 at 15:25
Deepesh Meena
4,21621025
i did not get how you got this $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
– JIM
Sep 4 at 15:39
I wrote it like that because I wanted to use $$mathrmdx=-dfrac1csc^2left(xright),mathrmdu$$ you can simply write that $$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ $$-1cdot-1=1$$
– Deepesh Meena
Sep 4 at 16:16
I also edited that the answer so you can understand what I am trying to do here
– Deepesh Meena
Sep 4 at 16:20
no I mean how did you get this step $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ from this step $$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$ ?
– JIM
Sep 4 at 16:21
take $sin(x)$ common from $$sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright)$$ $$sin^1/2(x)sqrtcosleft(aright)+sinleft(aright)cotleft(xright)$$
– Deepesh Meena
Sep 4 at 16:22
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
So you want to find another method here is one
$$I=displaystyleintdfrac1sin^frac32left(xright)sqrtsinleft(x+aright),mathrmdx$$
$$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$
$$sinleft(xright)=dfrac1cscleft(xright)$$
$$cosleft(xright)=dfraccotleft(xright)cscleft(xright)$$
$$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
$$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
Substitute $u=cotleft(xright)$ Thus $mathrmdx=-dfrac1csc^2left(xright),mathrmdu$
$$I=-displaystyleintdfrac1sqrtsinleft(aright)u+cosleft(aright),mathrmdu$$
$$v=sinleft(aright)u+cosleft(aright)$$
$$mathrmdu=dfrac1sinleft(aright),mathrmdv$$
$$I=-classsteps-nodecssIdsteps-node-3dfrac1sinleft(aright)displaystyleintdfrac1sqrtv,mathrmdv$$
$$I=-dfrac2sqrtvsinleft(aright)$$
$$I=-dfrac2sqrtsinleft(aright)u+cosleft(aright)sinleft(aright)$$
$$u=cotleft(xright)$$
$$I=-dfrac2sqrtsinleft(aright)cotleft(xright)+cosleft(aright)sinleft(aright)$$
$$I=-dfrac2sqrtsinleft(aright)cotleft(xright)+cosleft(aright)sinleft(aright)+C$$
answered Sep 4 at 15:25
Deepesh Meena
4,21621025
i did not get how you got this $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
– JIM
Sep 4 at 15:39
I wrote it like that because I wanted to use $$mathrmdx=-dfrac1csc^2left(xright),mathrmdu$$ you can simply write that $$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ $$-1cdot-1=1$$
– Deepesh Meena
Sep 4 at 16:16
I also edited that the answer so you can understand what I am trying to do here
– Deepesh Meena
Sep 4 at 16:20
no I mean how did you get this step $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ from this step $$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$ ?
– JIM
Sep 4 at 16:21
take $sin(x)$ common from $$sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright)$$ $$sin^1/2(x)sqrtcosleft(aright)+sinleft(aright)cotleft(xright)$$
– Deepesh Meena
Sep 4 at 16:22
 |Â
show 3 more comments
up vote
5
down vote
So you want to find another method here is one
$$I=displaystyleintdfrac1sin^frac32left(xright)sqrtsinleft(x+aright),mathrmdx$$
$$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$
$$sinleft(xright)=dfrac1cscleft(xright)$$
$$cosleft(xright)=dfraccotleft(xright)cscleft(xright)$$
$$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
$$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
Substitute $u=cotleft(xright)$ Thus $mathrmdx=-dfrac1csc^2left(xright),mathrmdu$
$$I=-displaystyleintdfrac1sqrtsinleft(aright)u+cosleft(aright),mathrmdu$$
$$v=sinleft(aright)u+cosleft(aright)$$
$$mathrmdu=dfrac1sinleft(aright),mathrmdv$$
$$I=-classsteps-nodecssIdsteps-node-3dfrac1sinleft(aright)displaystyleintdfrac1sqrtv,mathrmdv$$
$$I=-dfrac2sqrtvsinleft(aright)$$
$$I=-dfrac2sqrtsinleft(aright)u+cosleft(aright)sinleft(aright)$$
$$u=cotleft(xright)$$
$$I=-dfrac2sqrtsinleft(aright)cotleft(xright)+cosleft(aright)sinleft(aright)$$
$$I=-dfrac2sqrtsinleft(aright)cotleft(xright)+cosleft(aright)sinleft(aright)+C$$
answered Sep 4 at 15:25
Deepesh Meena
4,21621025
i did not get how you got this $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
– JIM
Sep 4 at 15:39
I wrote it like that because I wanted to use $$mathrmdx=-dfrac1csc^2left(xright),mathrmdu$$ you can simply write that $$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ $$-1cdot-1=1$$
– Deepesh Meena
Sep 4 at 16:16
I also edited that the answer so you can understand what I am trying to do here
– Deepesh Meena
Sep 4 at 16:20
no I mean how did you get this step $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ from this step $$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$ ?
– JIM
Sep 4 at 16:21
take $sin(x)$ common from $$sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright)$$ $$sin^1/2(x)sqrtcosleft(aright)+sinleft(aright)cotleft(xright)$$
– Deepesh Meena
Sep 4 at 16:22
 |Â
show 3 more comments
up vote
5
down vote
up vote
5
down vote
So you want to find another method here is one
$$I=displaystyleintdfrac1sin^frac32left(xright)sqrtsinleft(x+aright),mathrmdx$$
$$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$
$$sinleft(xright)=dfrac1cscleft(xright)$$
$$cosleft(xright)=dfraccotleft(xright)cscleft(xright)$$
$$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
$$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
Substitute $u=cotleft(xright)$ Thus $mathrmdx=-dfrac1csc^2left(xright),mathrmdu$
$$I=-displaystyleintdfrac1sqrtsinleft(aright)u+cosleft(aright),mathrmdu$$
$$v=sinleft(aright)u+cosleft(aright)$$
$$mathrmdu=dfrac1sinleft(aright),mathrmdv$$
$$I=-classsteps-nodecssIdsteps-node-3dfrac1sinleft(aright)displaystyleintdfrac1sqrtv,mathrmdv$$
$$I=-dfrac2sqrtvsinleft(aright)$$
$$I=-dfrac2sqrtsinleft(aright)u+cosleft(aright)sinleft(aright)$$
$$u=cotleft(xright)$$
$$I=-dfrac2sqrtsinleft(aright)cotleft(xright)+cosleft(aright)sinleft(aright)$$
$$I=-dfrac2sqrtsinleft(aright)cotleft(xright)+cosleft(aright)sinleft(aright)+C$$
answered Sep 4 at 15:25
Deepesh Meena
4,21621025
So you want to find another method here is one
$$I=displaystyleintdfrac1sin^frac32left(xright)sqrtsinleft(x+aright),mathrmdx$$
$$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$
$$sinleft(xright)=dfrac1cscleft(xright)$$
$$cosleft(xright)=dfraccotleft(xright)cscleft(xright)$$
$$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
$$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
Substitute $u=cotleft(xright)$ Thus $mathrmdx=-dfrac1csc^2left(xright),mathrmdu$
$$I=-displaystyleintdfrac1sqrtsinleft(aright)u+cosleft(aright),mathrmdu$$
$$v=sinleft(aright)u+cosleft(aright)$$
$$mathrmdu=dfrac1sinleft(aright),mathrmdv$$
$$I=-classsteps-nodecssIdsteps-node-3dfrac1sinleft(aright)displaystyleintdfrac1sqrtv,mathrmdv$$
$$I=-dfrac2sqrtvsinleft(aright)$$
$$I=-dfrac2sqrtsinleft(aright)u+cosleft(aright)sinleft(aright)$$
$$u=cotleft(xright)$$
$$I=-dfrac2sqrtsinleft(aright)cotleft(xright)+cosleft(aright)sinleft(aright)$$
$$I=-dfrac2sqrtsinleft(aright)cotleft(xright)+cosleft(aright)sinleft(aright)+C$$
answered Sep 4 at 15:25
Deepesh Meena
4,21621025
edited Sep 4 at 16:20
answered Sep 4 at 15:25
Deepesh Meena
4,21621025
answered Sep 4 at 15:25
Deepesh Meena
4,21621025
answered Sep 4 at 15:25
Deepesh Meena
4,21621025
4,21621025
i did not get how you got this $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
– JIM
Sep 4 at 15:39
I wrote it like that because I wanted to use $$mathrmdx=-dfrac1csc^2left(xright),mathrmdu$$ you can simply write that $$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ $$-1cdot-1=1$$
– Deepesh Meena
Sep 4 at 16:16
I also edited that the answer so you can understand what I am trying to do here
– Deepesh Meena
Sep 4 at 16:20
no I mean how did you get this step $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ from this step $$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$ ?
– JIM
Sep 4 at 16:21
take $sin(x)$ common from $$sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright)$$ $$sin^1/2(x)sqrtcosleft(aright)+sinleft(aright)cotleft(xright)$$
– Deepesh Meena
Sep 4 at 16:22
 |Â
show 3 more comments
i did not get how you got this $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
– JIM
Sep 4 at 15:39
I wrote it like that because I wanted to use $$mathrmdx=-dfrac1csc^2left(xright),mathrmdu$$ you can simply write that $$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ $$-1cdot-1=1$$
– Deepesh Meena
Sep 4 at 16:16
I also edited that the answer so you can understand what I am trying to do here
– Deepesh Meena
Sep 4 at 16:20
no I mean how did you get this step $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ from this step $$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$ ?
– JIM
Sep 4 at 16:21
take $sin(x)$ common from $$sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright)$$ $$sin^1/2(x)sqrtcosleft(aright)+sinleft(aright)cotleft(xright)$$
– Deepesh Meena
Sep 4 at 16:22
i did not get how you got this $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
– JIM
Sep 4 at 15:39
i did not get how you got this $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$
– JIM
Sep 4 at 15:39
I wrote it like that because I wanted to use $$mathrmdx=-dfrac1csc^2left(xright),mathrmdu$$ you can simply write that $$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ $$-1cdot-1=1$$
– Deepesh Meena
Sep 4 at 16:16
I wrote it like that because I wanted to use $$mathrmdx=-dfrac1csc^2left(xright),mathrmdu$$ you can simply write that $$I=displaystyleintclasssteps-nodecssIdsteps-node-1left(dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ $$-1cdot-1=1$$
– Deepesh Meena
Sep 4 at 16:16
I also edited that the answer so you can understand what I am trying to do here
– Deepesh Meena
Sep 4 at 16:20
I also edited that the answer so you can understand what I am trying to do here
– Deepesh Meena
Sep 4 at 16:20
no I mean how did you get this step $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ from this step $$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$ ?
– JIM
Sep 4 at 16:21
no I mean how did you get this step $$I=displaystyleint-classsteps-nodecssIdsteps-node-1left(-dfrac1sqrtsinleft(aright)cotleft(xright)+cosleft(aright)right)classsteps-nodecssIdsteps-node-2csc^2left(xright),mathrmdx$$ from this step $$I=displaystyleintdfrac1sin^frac32left(xright)sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright),mathrmdx$$ ?
– JIM
Sep 4 at 16:21
take $sin(x)$ common from $$sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright)$$ $$sin^1/2(x)sqrtcosleft(aright)+sinleft(aright)cotleft(xright)$$
– Deepesh Meena
Sep 4 at 16:22
take $sin(x)$ common from $$sqrtcosleft(aright)sinleft(xright)+sinleft(aright)cosleft(xright)$$ $$sin^1/2(x)sqrtcosleft(aright)+sinleft(aright)cotleft(xright)$$
– Deepesh Meena
Sep 4 at 16:22
 |Â
show 3 more comments
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Divide numerator & denominator by $cos^2x$ or $sin^2x$
– lab bhattacharjee
Sep 4 at 13:30
@labbhattacharjee I wanted to know if there was any other method for solving this or not ??
– JIM
Sep 4 at 13:42