If ∫$$x71$$−$$x25dx=1$$λ$$x81$$−$$x24+C$$ then λ is equal to

we have,$$I=$$∫$$x71$$−$$x25dx=$$∫$$sin7$$⁡θ$$1$$−$$sin2$$⁡θ$$5cos$$⁡θdθ, where $$x=$$$$sin$$⁡θ$$=$$∫$$tan7$$⁡θ$$sec2$$⁡θdθ$$=tan8$$⁡θ$$8+C=18x81$$−$$x24+C$$λ$$=8$$

Methods of integration

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Question

If $\int \frac{x^{7}}{{(1 - x^{2})}^{5}} dx = \frac{1}{λ} \frac{x^{8}}{{(1 - x^{2})}^{4}} + C$ then $λ$ is equal to

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Solution

we have,
$\begin{array}{l} I = \int \frac{x^{7}}{{(1 - x^{2})}^{5}} dx \\ = \int \frac{\sin^{7} θ}{{(1 - \sin^{2} θ)}^{5}} \cos θdθ, where x = \sin θ \\ = \int \tan^{7} {θsec}^{2} θdθ = \frac{\tan^{8} θ}{8} + C = \frac{1}{8} \frac{x^{8}}{{(1 - x^{2})}^{4}} + C \\ λ = 8 \end{array}$

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