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Question Number 176570 by mathlove last updated on 21/Sep/22

$$\left(1\right){\underset{\frac{\pi }{3}}{\int }}^{\frac{\pi }{2}}\frac{1+sinx}{cosx}dx=?$$

Answered by Peace last updated on 21/Sep/22

$$\int \frac{1+sin\left(x\right)}{cos\left(x\right)}dx=\int \frac{cos\left(x\right)}{{cos}^2\left(x\right)}+\int \frac{sin\left(x\right)}{cos\left(x\right)}dx$$$$=\int \frac{cos\left(x\right)}{1-{sin}^2\left(x\right)}dx-\int \frac{d\left(cos\left(x\right)\right)}{cos\left(x\right)}$$$$=\int \frac{d\left(sinx\right)}{1-{sin}^2\left(x\right)}-ln\mid cos\left(x\right)\mid$$$$\underset{}{=}{argth}^{-}\left(sin\left(x\right)\right)-ln\mid cos\left(x\right)\mid +c$$

Answered by BaliramKumar last updated on 23/Sep/22

$$\text{Missing left or extra right}$$$$=1+ln\left(2\right)-\frac{1}{\sqrt{3}}+2ln\left(\frac{\sqrt{3}}{2}\right)$$$$=1-ln\left(2\right)-\frac{1}{\sqrt{3}}+ln\left(3\right)$$$$=1-\frac{1}{\sqrt{3}}+ln\left(\frac{3}{2}\right)$$

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