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प्रश्न
\[\int \sin^3 x \cos^6 x \text{ dx }\]
बेरीज
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उत्तर
∫ sin3 x . cos6 x dx
= ∫ sin2 x . cos6 x . sin x dx
= ∫ (1 – cos2 x) . cos6 x . sin x dx
Let cos x = t
⇒ –sin x dx = dt
Now, ∫ (1 – cos2 x) . cos6 x . sin x dx
= –∫ (1 – t2) . t6 dt
= ∫ (t2 – 1) t6 dt
= ∫ (t8 – t6) dt
\[= \frac{t^9}{9} - \frac{t^7}{7} + C\]
\[ = \frac{\cos^9 x}{9} - \frac{\cos^7 x}{7} + C\]
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