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प्रश्न
\[\int \cos^7 x \text{ dx } \]
बेरीज
उत्तर
∫ cos7 x dx
= ∫ cos6 x . cos x dx
= ∫ (cos2 x)3 cos x dx
= ∫ (1 – sin2 x)3 . cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ (1 – sin2 x)3.cos x dx
= ∫ (1 – t2)3 dt
= ∫ (1 – t6 – 3t2 + 3t4) dt
\[= \left[ t - \frac{t^7}{7} - \frac{3 t^3}{3} + \frac{3 t^5}{5} \right] + C\]
\[ = \sin x - \frac{1}{7} \sin^7 x - \sin^3 x + \frac{3}{5} \sin^5 x + C\]
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