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प्रश्न
\[\int \sin^7 x \text{ dx }\]
बेरीज
उत्तर
∫ sin7 x dx
= ∫ sin6 x . sin x dx
= ∫ (sin2 x)3 sin x dx
= ∫ (1 – cos2 x)3 sin x dx
Let cos x = t
⇒ –sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos2 x)3 sin x dx
= ∫ (1 – t2)3 . (–dt)
= –∫ (1 – t6 – 3t2 + 3t4) dt
\[= - \left[ t - \frac{t^7}{7} - t^3 + \frac{3 t^5}{5} \right] + C\]
\[ = - \left[ \cos x - \frac{\cos^7 x}{7} - \cos^3 x + \frac{3}{5} \cos^5 x \right] + C\]
\[ = - \cos x + \frac{1}{7} \cos^7 x + \cos^3 x - \frac{3}{5} \cos^5 x + C\]
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