Q111 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Mathematics

$$ \int \frac{\cos ^4 x}{\left(\sin ^2 x+\sin ^{-3} x \cos ^5 x\right)^3} d x= $$

  • A
    $\frac{1}{5}\left(1+\cot ^5 x\right)^{-2}+C$
  • B
    $\frac{1}{10}\left(1+\cot ^2 x\right)^{-5}+C$
  • Correct answer
    C
    $\frac{1}{10}\left(1+\cot ^5 x\right)^{-2}+C$
  • D
    $\frac{1}{5}\left(1+\cot ^5 x\right)^{-5}+C$

Solution

$\begin{aligned} & \text { Let } I=\int \frac{\cos ^4 x}{\left(\sin ^2 x+\sin ^{-3} x \cos ^5 x\right)^3} d x \\ & =\int \frac{\cos ^4 x}{\left(\sin ^2 x\right)^3\left(1+\cot ^5 x\right)^3} \\ & =\int \frac{\cot ^4 x \operatorname{cosec}^2 x}{\left(1+\cot ^5 x\right)^3} d x \Rightarrow 1+\cot ^5 x=y \\ & \Rightarrow \quad 5 \cot ^4 x\left(-\operatorname{cosec}^2 x\right) d x=d y \\ & \Rightarrow \quad \cot ^4 x \operatorname{cosec}^2 x d x=\frac{-1}{5} d y \\ & I=\frac{-1}{5} \int \frac{1}{y^3} d y=\frac{-1}{5} \frac{y^{-2}}{-2} \\ & =\frac{1}{10} \times y^{-2}+C \\ & =\frac{1}{10}\left[1+\cot ^5 x\right]^{-2}+C\end{aligned}$

Hence, the correct answer is option: 'C'.

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