Q75 / 150 · MHT CET 2023 (12 May Shift 1) · Mathematics

The integral $\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x$ is equal to

  • A
    $\frac{1}{3\left(1+\tan ^3 x\right)}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • Correct answer
    B
    $\frac{-1}{3\left(1+\tan ^3 x\right)}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • C
    $\frac{1}{1+\cot ^3 x}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • D
    $\frac{-1}{1+\cos ^3 x}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.

Solution

Let $$ \begin{aligned} \mathrm{I} & =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x \\ & =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\sin ^3 x \cos ^2 x+\cos ^5 x \sin ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x \\ & =\int \frac{\sin ^2 x \cos ^2 x}{\left[\sin ^3 x\left(\sin ^2 x+\cos ^2 x\right)+\cos ^3 x\left(\sin ^2 x+\cos ^2 x\right)\right]^2} \mathrm{~d} x \\ & =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} \mathrm{~d} x \\ & =\int \frac{\sec ^2 x \tan ^2 x}{\left(1+\tan ^3 x\right)^2} \mathrm{~d} x \end{aligned} $$ [Dividing numerator and denominator by $\cos ^6 x$ ] Let $1+\tan ^3 x=t$ Differentiating w.r.t. $x$, we get $$ \begin{aligned} & 3 \tan ^2 x \sec ^2 x \mathrm{~d} x=\mathrm{dt} \\ & \tan ^2 x \sec ^2 x \mathrm{~d} x=\frac{1}{3} \mathrm{dt} \end{aligned} $$ $$ \therefore \quad \tan ^2 x \sec ^2 x \mathrm{~d} x=\frac{1}{3} \mathrm{dt} $$ $$ \begin{aligned} \therefore \quad \mathrm{I} & =\frac{1}{3} \int \frac{1}{\mathrm{t}^2} \mathrm{dt} \\ & =\frac{-1}{3 \mathrm{t}}+\mathrm{c} \\ & =\frac{-1}{3\left(1+\tan ^3 x\right)}+\mathrm{c} \end{aligned} $$

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

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