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#Integrals

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Find Value of
$\int \dfrac{1}{ \sqrt{ {sin}^{3} x \: sin(x + \alpha )} } dx \\ \\ \alpha \ne n\pi \: \: \: \: n = integer$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is .

$\red{\rm :\longmapsto\: \displaystyle\int \dfrac{1}{ \sqrt{ {sin}^{3} x \: sin(x + \alpha )} } dx}$

We know,

$\blue{ \boxed{\bf \:sin(x + y) = sinxcosy + sinycosx}}$

Using this, we get

$\rm \: = \: \displaystyle\int \dfrac{1}{ \sqrt{ {sin}^{3} x \: (sinxcos\alpha + sin \alpha cosx )} } dx$

Take out sinx common from denominator, we get

$\rm \: = \: \displaystyle\int \dfrac{1}{ \sqrt{ {sin}^{4} x \: (cos\alpha + sin \alpha \dfrac{cosx}{sinx} )} } dx$

$\rm \: = \: \displaystyle\int \dfrac{1}{ \sqrt{ {sin}^{4} x \: (cos\alpha + sin \alpha cotx )} } dx$

$\rm \: = \: \displaystyle\int \dfrac{1}{ {sin}^{2}x \sqrt{ (cos\alpha + sin \alpha cotx )} } dx$

$\rm \: = \: \displaystyle\int \dfrac{ {cosec}^{2} x}{ \sqrt{ (cos\alpha + sin \alpha cotx )} } dx$

Now, we use method of Substitution, to evaluate this integral further.

$\red{\rm :\longmapsto\:Put \: cos \alpha + sin \alpha cotx = y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto\:\dfrac{d}{dx}(cos \alpha + sin \alpha cotx) = \dfrac{d}{dx}y}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cos \alpha + \dfrac{d}{dx}sin \alpha cotx= \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\:sin \alpha \dfrac{d}{dx}cotx= \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: - sin \alpha \: {cosec}^{2}x = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: {cosec}^{2}x \: dx = - \dfrac{dy}{sin \alpha }}$

$\red{\rm :\longmapsto\: {cosec}^{2}x \: dx = -cosec \alpha \: {dy}}$

So, on substituting these values, in above integral,

$\rm \: = - \:cosec \alpha \displaystyle\int \dfrac{ 1}{ \sqrt{ y} } dy$

$\rm \: = - \:cosec \alpha \displaystyle\int {\bigg(y\bigg) }^{ - \dfrac{1}{2} } dy$

$\rm \: = \: \: - cosec \alpha \: \dfrac{ {\bigg(y\bigg) }^{ - \dfrac{1}{2} + 1 } }{ - \dfrac{1}{2} + 1} + c$

$\rm \: = \: \: - cosec \alpha \: \dfrac{ {\bigg(y\bigg) }^{ \dfrac{ - 1 + 2}{2}} }{ \dfrac{ - 1 + 2}{2}} + c$

$\rm \: = \: \: - cosec \alpha \: \dfrac{ {\bigg(y\bigg) }^{ \dfrac{1}{2}} }{ \dfrac{1}{2}} + c$

$\rm \: = \: \: - 2 \: cosec \alpha \: \sqrt{y} + c$

$\rm \: = \: \: - 2 \: cosec \alpha \: \sqrt{cos \alpha + sin \alpha \: cotx} + c$

$\rm \: = \: \: - 2 \: cosec \alpha \: \sqrt{cos \alpha + sin \alpha \: \dfrac{cosx}{sinx} } + c$

$\rm \: = \: \: - 2 \: cosec \alpha \: \sqrt{ \dfrac{sinx \: cos \alpha \: + sin \alpha \: cosx}{sinx} } + c$

$\rm \: = \: \: - 2 \: cosec \alpha \: \sqrt{ \dfrac{sin(x \: + \: \alpha)}{sinx} } + c$

Hence,

The solution of

$\red{\rm :\longmapsto\: \displaystyle\int \bf \: \dfrac{1}{ \sqrt{ {sin}^{3} x \: sin(x + \alpha )} } dx}$

$\bf \: = \: \: - 2 \: cosec \alpha \: \sqrt{ \dfrac{sin(x \: + \: \alpha)}{sinx} } + c$

Additional Information :-

$\blue{ \boxed{\displaystyle\int\bf kdx = kx + c}}$

$\blue{ \boxed{\displaystyle\int\bf {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1} + c}}$

$\blue{ \boxed{\displaystyle\int\bf {e}^{x} dx = {e}^{x} + c}}$

$\blue{ \boxed{\displaystyle\int\bf {a}^{x} dx = \frac{ {a}^{x} }{loga} + c}}$

$\blue{ \boxed{\displaystyle\int\bf cosx \: dx = sinx + c}}$

$\blue{ \boxed{\displaystyle\int\bf sinx \: dx = - \: cosx + c}}$

$\blue{ \boxed{\displaystyle\int\bf cosecx \: cotx \: dx = - \: cosecx + c}}$

$\blue{ \boxed{\displaystyle\int\bf secx \: tanx \: dx = \: secx + c}}$

Answer 2

Answer:

hope it helps you