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Question Number 51501 by Tinkutara last updated on 27/Dec/18

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18

$$x=tan\theta dx={sec}^2\theta d\theta$$$$\int \frac{\theta }{{tan}^4\theta }{sec}^2\theta d\theta$$$$\theta \int \frac{{sec}^2\theta }{{tan}^4\theta }d\theta -\int [\frac{d\theta }{d\theta }\int \frac{{sec}^2\theta }{{tan}^4\theta }d\theta ]d\theta$$$$=\theta \times \frac{-1}{3{tan}^3\theta }+\frac{1}{3}\int \frac{d\theta }{{tan}^3\theta }$$$$now\frac{1}{3}\int \frac{d\theta }{{tan}^3\theta }$$$$\frac{1}{3}\int \frac{(1-{sin}^2\theta )cos\theta }{{sin}^3\theta }d\theta$$$$\frac{1}{3}\int (\frac{1}{{sin}^3\theta }-\frac{1}{sin\theta })d\left(sin\theta \right)$$$$=\frac{1}{3}[\frac{{\left(sin\theta \right)}^{-3+1}}{-3+1}-ln\left(sin\theta \right)]$$$$answeris$$$$\frac{-\theta }{3{tan}^3\theta }+\frac{-1}{6}\times \frac{1}{{sin}^2\theta }-\frac{1}{3}ln\left(sin\theta \right)+c$$$$=\frac{-{tan}^{-1}x}{3x^3}+\frac{-1}{6}\times \frac{1+x^2}{x^2}-\frac{1}{3}ln\left(\frac{x}{\sqrt{1+x^2}}\right)+c$$$$plscheck$$$$$$$$[\int \frac{{sec}^{2}d\theta }{{tan}^4\theta }=\frac{{\left(tan\theta \right)}^{-4+1}}{-4+1}+c]$$

Commented by Tinkutara last updated on 27/Dec/18

Thank you very much Sir! I got the answer. ��������

Commented by Tinkutara last updated on 27/Dec/18

$$\frac{-1}{6}\times \frac{1+x^2}{x^2}=\frac{-1}{6x^2}+c$$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18

$$mostwelcome...$$

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