Q113 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics

$\int \sin ^4 x \cos ^4 x d x=$

  • A
    $\frac{1}{128}\left(-2 \sin ^3 x \cos x-3 \sin x \cos x+3\right)+c$
  • Correct answer
    B
    $\frac{1}{256}\left(-2 \sin ^3 2 x \cos 2 x-3 \sin 2 x \cos 2 x+6 x\right)+c$
  • C
    $\frac{1}{128}\left(2 \sin ^3 x \cos x-3 \sin x \cos x+3 x\right)+c$
  • D
    $\frac{1}{256}\left(3 \sin ^3 x \cos x-2 \sin x \cos x+2\right)+c$

Solution

$\begin{aligned} & \text { } \mathrm{I}=\int \sin ^4 x \cos ^4 x d x=\frac{1}{16} \int(\sin 2 x)^4 d x \\ & \mathrm{I}=\frac{1}{16} \int \frac{[1-\cos 4 x]^2}{4} d x=\frac{1}{64} \int\left[1+\cos ^2 4 x-2 \cos 4 x\right] d x \\ & \mathrm{I}=\frac{1}{64} \int\left[1+\left(\frac{1+\cos 8 x}{2}\right)-2 \cos 4 x\right] d x \\ & \mathrm{I}=\frac{1}{128} \int(2+1+\cos 8 x-4 \cos 4 x) d x \\ & \mathrm{I}=\frac{1}{128}\left[3 x+\frac{\sin 8 x}{8}-\frac{4 \sin 4 x}{4}\right]+c \\ & \mathrm{I}=\frac{1}{128}\left[3 x+\frac{2 \sin 4 x \cos 4 x}{8}-\sin 4 x\right]+c \\ & \mathrm{I}=\frac{1}{128}\left[3 x+\frac{2 \sin 2 x \cos 2 x}{4}\left\{\left(1-2 \sin ^2 2 x\right)-4\right\}\right]+c \\ & \mathrm{I}=\frac{1}{256}\left[6 x-2 \sin ^3 2 x \cos 2 x-3 \sin 2 x \cos 2 x\right]+\mathrm{c}\end{aligned}$

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

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