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Question Number 36459 by rahul 19 last updated on 02/Jun/18

$$\int {\mathrm{e}}^{\tan \theta }(\sec \theta -\sin \theta )\mathrm{d}\theta =?$$

Answered by ajfour last updated on 02/Jun/18

$$let\tan \theta =t$$$$\Rightarrow I=\int e^t(\sec \theta -\sin \theta )\cos {}^2\theta dt$$$$=\int e^t(\cos \theta -\frac{\sin \theta }{1+\tan {}^2\theta })dt$$$$=\int e^t(\frac{1}{\sqrt{1+t^2}}-\frac{t}{{(1+t^2)}^{3/2}})dt$$$$=\int e^t[\frac{1}{\sqrt{1+t^2}}+\frac{d}{dt}\left(\frac{1}{\sqrt{1+t^2}}\right)]dt$$$$=\frac{e^t}{\sqrt{1+t^2}}+c$$$$I=e^{\tan \theta }\cos \theta +c.$$

Commented by rahul 19 last updated on 02/Jun/18

Thank you sir ����

Answered by MJS last updated on 02/Jun/18

$$\mathrm{lol},\mathrm{tricky}...$$$$\int {\mathrm{e}}^{\tan \theta }(\sec \theta -\sin \theta )d\theta =$$$$=\int {\mathrm{e}}^{\tan \theta }\sec \theta d\theta -\int {\mathrm{e}}^{\tan \theta }\sin \theta d\theta =$$$$\left[\begin{array}{c}{f}^{\prime }=\sin \theta \Rightarrow f=-\cos \theta \\ g={\mathrm{e}}^{\tan \theta }\Rightarrow g^{\prime }={\mathrm{e}}^{\tan \theta }{\sec }^2\theta \\ \int f^{\prime }g=fg-\int {fg}^{\prime }\end{array}\right]$$$$=\int {\mathrm{e}}^{\tan \theta }\sec \theta d\theta +{\mathrm{e}}^{\tan \theta }\cos \theta -\int {\mathrm{e}}^{\tan \theta }\cos \theta {\sec }^2\theta d\theta =$$$$={\mathrm{e}}^{\tan \theta }\cos \theta +C$$

Commented by rahul 19 last updated on 02/Jun/18

$$\mathrm{How}\mathrm{fg}=\int {\mathrm{f}}^{{}^{\prime }}\mathrm{g}+\int {\mathrm{fg}}^{\prime }??$$$$\mathrm{Integration}\mathrm{by}\mathrm{parts}\mathrm{says},$$$$\int \mathrm{fg}=\mathrm{f}\int \mathrm{g}-\int {\mathrm{f}}^{\prime }\int \mathrm{g}.$$

Commented by MJS last updated on 02/Jun/18

$$https://en.wikipedia.org/wiki/Integration\mathrm{\_}by\mathrm{\_}parts$$

Commented by rahul 19 last updated on 02/Jun/18

$$\mathrm{Sir}\mathrm{if}\mathrm{you}\mathrm{want}\mathrm{to}\mathrm{integrate}\mathrm{say}$$$$\int x\sin x\mathrm{then}\mathrm{g}=\sin x,\mathrm{f}=x.\left(\mathrm{ILATE}\right)$$$$\Rightarrow \int \mathrm{g}=-\cos x,{\mathrm{f}}^{\prime }=1.$$$$\int \mathrm{fg}=\mathrm{f}\int \mathrm{g}-\int {\mathrm{f}}^{\prime }\int \mathrm{g}$$$$\Rightarrow x(-\cos x)-\int -\left(\cos x\right)$$$$\Rightarrow -x\cos x+\sin x+c.$$

Commented by MJS last updated on 02/Jun/18

$$\mathrm{we}\mathrm{come}\mathrm{from}\mathrm{differentation}:$$$${\left(uv\right)}^{\prime }=u^{\prime }v+{uv}^{\prime }$$$$\mathrm{now}\mathrm{integrate}\mathrm{both}\mathrm{sides}:$$$$uv=\int u^{\prime }v+\int {uv}^{\prime }$$$${\mathrm{that}}^{\prime }\mathrm{s}\mathrm{how}\mathrm{I}\mathrm{learned}\mathrm{it}...$$

Commented by rahul 19 last updated on 02/Jun/18

$$\mathrm{Is}\mathrm{this}\mathrm{also}\mathrm{called}\mathrm{integration}\mathrm{by}\mathrm{parts}?$$$$\mathrm{Anyways},\mathrm{thanks}!!$$

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