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

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

Grade:12

1 Answers

Arun
25750 Points
4 years ago
I = \frac{\pi }{12}2I = \frac{\pi }{6}2I = [x]_{\pi /6}^{\pi /3}2I =\int_{\pi /6}^{\pi /3}dx2I =\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)}}dxI =\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}}dxI =\int_{\pi /6}^{\pi /3}\frac{1}{1+\sqrt{tanx}}dxHello student, please find answer to your question
 
Hope it helps
 
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