TS EAMCET · Maths · Indefinite Integration
If \(\mathrm{I}_1=\int \sin ^6 x d x\) and \(\mathrm{I}_2=\int \cos ^6 x d x\) then \(\mathrm{I}_1+\mathrm{I}_2=\)
- A \(\frac{5 x}{8}+\frac{3 \cos 4 x}{32}+c\)
- B \(\frac{1}{32}(20 x-3 \sin 4 x)+c\)
- C \(\frac{1}{32}(20 x+3 \sin 4 x)+c\)
- D \(\frac{5 x}{4}+\frac{3 \sin 4 x}{16}+c\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{32}(20 x+3 \sin 4 x)+c\)
Step-by-step Solution
Detailed explanation
\( \mathrm{I}_1+\mathrm{I}_2=\int (\sin ^6 x+\cos ^6 x) d x \) \( \sin ^6 x+\cos ^6 x = (\sin ^2 x+\cos ^2 x)(\sin ^4 x-\sin ^2 x \cos ^2 x+\cos ^4 x) \) \( = 1 \cdot ((\sin ^2 x+\cos ^2 x)^2-3 \sin ^2 x \cos ^2 x) \) \( = 1 - 3 \sin ^2 x \cos ^2 x \)…
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