Question

Solve the following integral

$\int \: \frac{dx}{ \sqrt{ {sin}^{3} x \: sin(x + a)} }$
​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \: \frac{dx}{ \sqrt{ {sin}^{3}x \: sin(x + a)} }$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {sin}^{3}x(sinx \: cosa + sina \: cosx )} }$

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {sin}^{3}x\bigg(sinx \: cosa + sina \:sinx \times \dfrac{cosx}{sinx} \bigg)} }$

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {sin}^{4}x\bigg(\: cosa + sina \:cotx \bigg)} }$

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{ {sin}^{2} x \sqrt{ \: cosa + sina \:cotx } }$

$\rm \:  =  \: \displaystyle\int\rm \frac{ {cosec}^{2}x \: dx}{\sqrt{ \: cosa + sina \:cotx } }$

Now, to evaluate this integral, we use Method of Substitution.

So, Substitute

$\purple{\rm :\longmapsto\: \sqrt{cosa + sina \: cotx }\: = \: y}$

$\purple{\rm :\longmapsto\:cosa + sina \: cotx \: = \: {y}^{2} }$

$\purple{\rm :\longmapsto\: - \: sina \: {cosec}^{2} x \: dx \: = \: 2y \: dy }$

$\purple{\rm :\longmapsto\: \: {cosec}^{2} x \: dx \: = \: - \: \dfrac{2y}{sina} \: dy }$

$\purple{\rm :\longmapsto\: \: {cosec}^{2} x \: dx \: = \: - \: 2y \: coseca \: \: dy }$

So, on substituting the values, we get

$\rm \:  =  \: \displaystyle\int\rm \frac{ - 2y \: coseca \: }{y} \: dy$

$\rm \:  =  \: - \: 2 \: coseca \: \displaystyle\int\rm dy$

$\rm \:  =  \: - \: 2 \: coseca \:y \: + \: c$

$\rm \:  =  \: - \: 2 \: coseca \:( \sqrt{cosa + sina \: cotx} \: + \: c$

Hence,

$\boxed{\tt{ \rm \displaystyle\int\rm \frac{dx}{ \sqrt{ {sin}^{3}x \: sin(x + a)} } =- 2 coseca \: ( \sqrt{cosa + sina \: cotx} + \: c}}$

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MORE TO KNOW

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \: \frac{dx}{ \sqrt{ {sin}^{3}x \: sin(x + a)} } \\ \\ can \: \: be \: \: rewritten \: \: as \\ \\ \rm \: = \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {sin}^{3}x(sinx \: cosa + sina \: cosx )} } \\ \\ \rm \: = \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {sin}^{3}x\bigg(sinx \: cosa + sina \:sinx \times \dfrac{cosx}{sinx} \bigg)} } \\ \\ \rm \: = \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {sin}^{4}x\bigg(\: cosa + sina \:cotx \bigg)} } \\ \\ \rm \: = \: \displaystyle\int\rm \frac{dx}{ {sin}^{2} x \sqrt{ \: cosa + sina \:cotx } }$

$\rm \: = \: \displaystyle\int\rm \frac{ {cosec}^{2}x \: dx}{\sqrt{ \: cosa + sina \:cotx } }$

Now, to evaluate this integral, we use Method of Substitution.

So, Substitute

$\purple{\rm :\longmapsto\: \sqrt{cosa + sina \: cotx }\: = \: y} \\ \\ \purple{\rm :\longmapsto\:cosa + sina \: cotx \: = \: {y}^{2} } \\ \\ \purple{\rm :\longmapsto\: - \: sina \: {cosec}^{2} x \: dx \: = \: 2y \: dy } \\ \\ \purple{\rm :\longmapsto\: \: {cosec}^{2} x \: dx \: = \: - \: \dfrac{2y}{sina} \: dy } \\ \\ \purple{\rm :\longmapsto\: \: {cosec}^{2} x \: dx \: = \: - \: 2y \: coseca \: \: dy }$

So, on substituting the values, we get

$\rm \: = \: \displaystyle\int\rm \frac{ - 2y \: coseca \: }{y} \: dy \\ \\ \rm \: = \: - \: 2 \: coseca \: \displaystyle\int\rm dy \\ \\ \rm \: = \: - \: 2 \: coseca \:y \: + \: c \\ \\ \rm \: = \: - \: 2 \: coseca \:( \sqrt{cosa + sina \: cotx} \: + \: c \\ \\ Hence, \\ \\ \boxed{\tt{ \rm \displaystyle\int\rm \frac{dx}{ \sqrt{ {sin}^{3}x \: sin(x + a)} } =- 2 coseca \: ( \sqrt{cosa + sina \: cotx} + \: c}}$

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