Q115 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x=$

  • A
    $\frac{2}{3}$
  • B
    $\frac{3}{10}$
  • Correct answer
    C
    $\frac{4}{15}$
  • D
    $\frac{5}{18}$

Solution

$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x$...(i) Using $\int_a^b f(x)=\int_a^b f(a+b-x)$ $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cos ^2 x(-\sin x+\cos x) d x$...(ii) $\begin{aligned} & \text { Eq. (i) }+ \text { (ii) } \\ & 2 I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cos ^2 x(2 \cos x) d x \\ & I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cdot \cos ^3 x d x \\ & =2 \int_0^{\frac{\pi}{2}} \sin ^2 x \cdot \cos ^3 x d x \\ & \text { Let } \sin x=t \Rightarrow \cos x d x=d t \\ & =2 \int_0^1 t^2\left(1-t^2\right) d t \\ & =2 \int_0^1\left(t^2-t^4\right) d t=2\left[\frac{t^3}{3}-\frac{t^5}{5}\right]_0^1=\frac{4}{15} \\ & \end{aligned}$

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

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