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प्रश्न
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
बेरीज
उत्तर
∫ cosec4 3x dx
= ∫ cosec2 3x . cosec2 3x dx
= ∫ (1 + cot2 3x) cosec2 3x dx
Let cot (3x) = t
⇒ –cosec2 (3x) × 3 dx = dt
\[\Rightarrow {cosec}^2 \left( 3x \right)dx = - \frac{dt}{3}\]
\[Now, \int\left( 1 + \cot^2 3x \right)\]
\[ = \frac{- 1}{3}\int\left( 1 + t^2 \right) dt\]
\[ = \frac{- 1}{3} \left[ t + \frac{t^3}{3} \right] + C\]
\[ = - \frac{t}{3} - \frac{t^3}{9} + C\]
\[ = \frac{- \text{cot} \left( 3x \right)}{3} - \frac{\text{ cot }^3 3x}{9} + C\]
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