Question:
\(\int \sin^3 2x \sin^{26}x \, dx=\)
\(\int \sin^3 2x \sin^{26}x \, dx=\)
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Convert everything to single trig function before integrating.
Updated On: Jun 22, 2026
- \(8(\frac{\sin^{27}x}{27}-\frac{\sin^{29}x}{29})+c\)
- \(4(\frac{\sin^{28}x}{14}-\frac{\sin^{30}x}{15})+c\)
- \(8(\frac{\sin^{31}x}{31}-\frac{\sin^{33}x}{33})+c\)
- \(4(\frac{\sin^{30}x}{15}-\frac{\sin^{32}x}{16})+c\) \bigskip
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The Correct Option is C
Solution and Explanation
Concept:
Use:
\[
\sin 2x=2\sin x \cos x
\]
Step 1: Rewrite.
\[ \sin^3 2x=8\sin^3 x \cos^3 x \] \[ \Rightarrow 8\sin^{29}x \cos^3 x \]
Step 2: Substitute \(t=\sin x\).
\[ dt=\cos x dx \] Integral reduces to polynomial in \(t\)
Step 3: Final result.
\[ 8\left(\frac{\sin^{31}x}{31}-\frac{\sin^{33}x}{33}\right)+c \] \[ \boxed{(C)} \]
Step 1: Rewrite.
\[ \sin^3 2x=8\sin^3 x \cos^3 x \] \[ \Rightarrow 8\sin^{29}x \cos^3 x \]
Step 2: Substitute \(t=\sin x\).
\[ dt=\cos x dx \] Integral reduces to polynomial in \(t\)
Step 3: Final result.
\[ 8\left(\frac{\sin^{31}x}{31}-\frac{\sin^{33}x}{33}\right)+c \] \[ \boxed{(C)} \]
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