AP EAMCET · Maths · Indefinite Integration
\(\int \sin ^4 x \cos ^4 x d x=\)
- A \(\frac{1}{128}\left(-2 \sin ^3 x \cos x-3 \sin x \cos x+3\right)+c\)
- 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 \(\frac{1}{256}\left(3 \sin ^3 x \cos x-2 \sin x \cos x+2\right)+c\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{256}\left(-2 \sin ^3 2 x \cos 2 x-3 \sin 2 x \cos 2 x+6 x\right)+c\)
Step-by-step Solution
Detailed explanation
\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}…
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