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Intregation of dx/1+√tanx

Intregation of xdx/1+sinx
Intregation of (x+2) dx/2x^2+6x+5

Annesha biswas , 6 Years ago

Grade 12

1 Answers

Arun

$I = \frac{\pi }{12}$ $2I = \frac{\pi }{6}$ $2I = [x]_{\pi /6}^{\pi /3}$ $2I =\int_{\pi /6}^{\pi /3}dx$ $2I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{cosx}+\sqrt{sin(x)}}{\sqrt{sin(x)}+\sqrt{cos(x)}}dx$ ......(2)(1) + (2) $I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{sin(x)}}{\sqrt{sin(x)}+\sqrt{cos(x)}}dx$ $I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{cos(\frac{\pi }{2}-x)}}{\sqrt{cos(\frac{\pi }{2}-x)}+\sqrt{sin(\frac{\pi }{2}-x)}}dx$ $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ …......(1) $I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx$ $I =\int_{\pi /6}^{\pi /3}\frac{1}{1+\sqrt{tanx}}dx$ Hello student, please find answer to your question

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Last Activity: 6 Years ago