AP EAMCET · Maths · Indefinite Integration
\(\int \sin ^3 x \cos ^2 x d x=\)
- A \(\frac{\sin ^4 x \cos x}{5}-\frac{\sin ^2 x \cos x}{15}-\frac{2 \cos x}{15}+c\)
- B \(-\frac{\sin ^4 x \cos x}{5}-\frac{\sin ^2 x \cos x}{15}+\frac{2 \cos x}{15}+c\)
- C \(\frac{\sin ^4 x \cos x}{5}-\frac{\sin ^2 x \cos x}{15}+\frac{2 x}{15}+c\)
- D \(\frac{\sin ^4 x \cos x}{5}+\frac{\sin ^2 x \cos x}{3}-\frac{2 x}{15}+c\)
Answer & Solution
Correct Answer
(A) \(\frac{\sin ^4 x \cos x}{5}-\frac{\sin ^2 x \cos x}{15}-\frac{2 \cos x}{15}+c\)
Step-by-step Solution
Detailed explanation
\( \int \sin^3 x \cos^2 x dx = \int (1 - \cos^2 x) \cos^2 x \sin x dx \) \( = -\int (\cos^2 x - \cos^4 x) d(\cos x) \) \( = -\left( \frac{\cos^3 x}{3} - \frac{\cos^5 x}{5} \right) + c = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + c \)…
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