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Question Number 14394 by tawa tawa last updated on 31/May/17

$$\mathrm{Evaluate}\int \left(\frac{\mathrm{tanx}-\mathrm{cotx}}{\mathrm{tanx}+\mathrm{cotx}}{\sec }^2\mathrm{x}\right)\mathrm{dx}$$

Answered by RasheedSindhi last updated on 31/May/17

$$\mathrm{Evaluate}\int \left(\frac{\mathrm{tanx}-\mathrm{cotx}}{\mathrm{tanx}+\mathrm{cotx}}{\sec }^2\mathrm{x}\right)\mathrm{dx}$$$$\int \left(\frac{\frac{\mathrm{sinx}}{\mathrm{cosx}}-\frac{\mathrm{cosx}}{\mathrm{sinx}}}{\frac{\mathrm{sinx}}{\mathrm{cosx}}+\frac{\mathrm{cosx}}{\mathrm{sinx}}}{\sec }^2\mathrm{x}\right)\mathrm{dx}$$$$\int \left(\frac{{\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}}{{\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}}\times \frac{1}{{\cos }^2\mathrm{x}}\right)\mathrm{dx}$$$$\int \left(\frac{{\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}}{{\cos }^2\mathrm{x}}\right)\mathrm{dx}$$$$\int ({\tan }^2\mathrm{x}-1)\mathrm{dx}$$$$\int \left({\tan }^2\mathrm{x}\right)\mathrm{dx}-\int \left(1\right)\mathrm{dx}$$$$\int ({\sec }^2\mathrm{x}-1)\mathrm{dx}-\mathrm{x}+\mathrm{C}$$$$\mathrm{tanx}-\mathrm{x}-\mathrm{x}+\mathrm{C}$$$$\mathrm{tanx}-2\mathrm{x}+\mathrm{C}$$$$$$

Commented by tawa tawa last updated on 31/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by tawa tawa last updated on 31/May/17

$$\mathrm{Sir}\int {\sec }^2\mathrm{x}=\mathrm{tanx}+\mathrm{C}...\mathrm{mistake}$$

Commented by RasheedSindhi last updated on 31/May/17

$$\mathrm{Mistake}.\mathrm{Going}\mathrm{to}\mathrm{correct}.\mathrm{Thanks}!$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 31/May/17

$$u=tgx\Rightarrow du=(1+{tg}^{2}x)dx={sec}^{2}xdx$$$$I=\int \frac{u-\frac{1}{u}}{u+\frac{1}{u}}du=\int \frac{u^2-1}{u^2+1}du=$$$$=\int \frac{u^2+1-2}{u^2+1}du=\int (1-\frac{2}{u^2+1})du=$$$$=u-2{tg}^{-1}u+C=tgx-2x+\mathit{C}.◼$$

Commented by tawa tawa last updated on 31/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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