Q116 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics

$\int_0^\pi\left(\sin ^3 x+\cos ^2 x\right)^2 d x=$

  • A
    $\frac{15 \pi}{16}+\frac{8}{15}$
  • Correct answer
    B
    $\frac{11 \pi}{16}+\frac{8}{15}$
  • C
    $\frac{15 \pi}{16}+\frac{4}{15}$
  • D
    $\frac{11 \pi}{16}+\frac{4}{15}$

Solution

$\begin{aligned} & \text {} \quad I=\int_0^\pi\left(\sin ^3 x+\cos ^2 x\right)^2 d x \\ & =\int_0^\pi \sin ^6 x+\cos ^4 x+2 \cos ^2 x \sin ^3 x d x \\ & =2 \int_0^{\frac{\pi}{2}} \sin ^6 x d x+2 \int_0^{\frac{\pi}{2}} \cos ^4 x d x+\int_0^{\frac{\pi}{2}} \cos ^2 x \sin ^3 x d x \\ & =\frac{2 \times 5 \times 3 \times 1}{6 \times 4 \times 2} \times \frac{\pi}{2}+\frac{2 \times 3 \times 1}{4 \times 2} \times \frac{\pi}{2}+\frac{4 \times 1 \times 2}{5 \times 3} \\ & = \\ & =\frac{5 \pi}{16}+\frac{3 \pi}{8}+\frac{8}{15}=\frac{11 \pi}{16}+\frac{8}{15} .\end{aligned}$

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

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