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Question Number 140581 by bounhome last updated on 09/May/21

$$prove\int {sec}^{2}xdx=tanx$$

Answered by MJS_new last updated on 09/May/21

$$\tan x=\frac{\sin x}{\cos x}$$$$\frac{d}{dx}\left[\frac{u\left(x\right)}{v\left(x\right)}\right]=\frac{u^{\prime }\left(x\right)v\left(x\right)-u\left(x\right)v^{\prime }\left(x\right)}{{\left(v\left(x\right)\right)}^2}$$$$\Rightarrow$$$$\frac{d}{dx}\left[\frac{\sin x}{\cos x}\right]=\frac{\cos x\cos x-s\mathrm{in}x(-\sin x)}{{\cos }^{2}x}=$$$$=\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}={\sec }^{2}x$$$$\int \left(\frac{d}{dx}\left[f\left(x\right)\right]\right)dx=f\left(x\right)+C$$$$\Rightarrow$$$$\int {\sec }^{2}xdx=\tan x+C$$

Answered by MJS_new last updated on 10/May/21

$$\int {\sec }^{2}xdx=$$$$[t=\tan x\to dx={\cos }^{2}xdt]$$$$=\int dt=t=\tan x+C$$

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