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Question Number 43418 by Raj Singh last updated on 10/Sep/18

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

$$\int \frac{cosxdx}{sinxcosx+1}$$$$\int \frac{2cosx}{2cosxsinx+2}dx$$$$\int \frac{cosx+sinx+cosx-sinx}{1+{(sinx+cosx)}^2}dx$$$$\int \frac{cosx-sinx}{1+{(sinx+cosx)}^2}dx+\int \frac{cosx+sinx}{2+2sinxcosx}dx$$$$\int \frac{d(sinx+cosx)}{1+(sinx+cosx)}-\int \frac{(cosx+sinx)dx}{-3+1-2sinxcosx}$$$$\int \frac{d(sinx+cosx)}{1+{(sinx+cosx)}^2}-\int \frac{d(sinx-cosx)}{-3+{(sinx-cosx)}^2}$$$$formula\int \frac{dx}{x^2+a^2}=\frac{1}{a}{tan}^{-}\left(\frac{x}{a}\right)$$$$\int \frac{dx}{x^2-b^2}$$$$=\frac{1}{2b}\int \frac{(x+b)-(x-b)}{(x+b)(x-b)}dx$$$$=\frac{1}{2b}[\int \frac{dx}{x-b}-\int \frac{dx}{x+b}]$$$$=\frac{1}{2b}ln\mid \frac{x-b}{x+b}\mid$$$$soansis$$$${tan}^{-1}(sinx+cosx)-\frac{1}{2\sqrt{3}}ln\mid \frac{sinx-cosx-\sqrt{3}}{sinx-cosx+\sqrt{3}}\mid +c$$$$$$

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