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प्रश्न
\[\int \sin^5 x \text{ dx }\]
बेरीज
उत्तर
∫ sin5 x dx
= ∫ sin4 x . sin x dx
= ∫ (1 – cos2 x)2 sin x dx
= ∫ (1 – cos4 x – 2 cos2 x) sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos4 x – 2 cos2 x) sin x dx
=–∫ (1 + t4 – 2t2) dt
\[= - \left[ t + \frac{t^5}{5} - \frac{2 t^3}{3} \right] + C\]
\[ = - t - \frac{t^5}{5} + \frac{2 t^3}{3} + C\]
\[ = - \cos x + \frac{2}{3} \cos^3 x - \frac{\cos^5 x}{5} + C\]
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