Question

intergral of 1/(sinx+cosx)​

Answer 1

Question :

$\bullet\ \; \displaystyle \sf \red{\int \dfrac{1}{sin\ x+cos\ x}\ dx}$

Solution :

$\displaystyle \sf \int \dfrac{1}{sin\ x+cos\ x} \; dx$

We know that ,

$\bullet\ \; \sf \pink{sin\ x=\dfrac{2\ tan\ \frac{x}{2}}{1+tan^2 \frac{x}{2}}}$

$\bullet\ \; \sf \green{cos\ x=\dfrac{1-tan^2 \frac{x}{2}}{1+tan^2 \frac{x}{2}}}$

$\\ \to \displaystyle \sf \int \dfrac{1}{ \dfrac{2\ tan\ \frac{x}{2}}{1+tan^2 \frac{x}{2}} +\dfrac{1-tan^2 \frac{x}{2}}{1+tan^2 \frac{x}{2}}} \; dx$

$\\ \to \displaystyle \sf \int \dfrac{1+tan^2 \frac{x}{2}}{2\ tan\ \frac{x}{2}+1-tan^2 \frac{x}{2}}\ dx$

$\bullet\ \; \purple{\sf sec^2 \theta =1+tan^2 \theta }$

$\\ \to \displaystyle \sf \int \dfrac{sec^2 \frac{x}{2} }{2\ tan\ \frac{x}{2}+1-tan^2 \frac{x}{2}}\ dx$

$\\ \to \displaystyle \sf \int \dfrac{sec^2 \frac{x}{2} }{2\ tan\ \frac{x}{2}+2-1-tan^2 \frac{x}{2}}\ dx$

$\\ \to \displaystyle \sf \int \dfrac{sec^2 \frac{x}{2} }{2-(tan^2 \frac{x}{2}+1+2\ tan\ \frac{x}{2})}\ dx$

$\\ \to \displaystyle \sf \int \dfrac{sec^2 \frac{x}{2} }{(\sqrt{2})^2-\left(tan\ \frac{x}{2}+1 \right)^2}\ dx$

Let's use substitution method ,

➠ $\sf u=tan\ \frac{x}{2}+1$

➠ $\sf du=\frac{1}{2}\ sec^2\frac{x}{2}\ dx$

➠ $\sf 2\ du=sec^2 \frac{x}{2}\ dx$

$\\ \to \displaystyle \sf \int \dfrac{2\ du }{(\sqrt{2})^2-(u)^2}$

$\\ \to \displaystyle \sf 2\ \int \dfrac{ du }{(\sqrt{2})^2-(u)^2}$

$\bullet\ \; \displaystyle \sf \orange{\int \dfrac{dx}{a^2-x^2}=\dfrac{1}{2a}\ \ln \bigg| \dfrac{a+x}{a-x} \bigg| }$

$\to \sf 2\ \dfrac{1}{2 \sqrt{2}}\ \ln \bigg| \dfrac{\sqrt{2}+u}{\sqrt{2}-u} \bigg| + c$

$\to \sf \blue{ \dfrac{1}{ \sqrt{2}}\ \ln \bigg| \dfrac{\sqrt{2}+ tan\ \frac{x}{2}-1}{\sqrt{2}-tan\ \frac{x}{2}-1} \bigg| +c}\ \; \bigstar$

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Answer 2

The Requested Answer Of The

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