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Integral Calculus> Integrate ∫((tanx) 1/2 +(cotx) 1/2 ) dx...

question mark

Integrate
∫((tanx)^1/2+(cotx)^1/2) dx

Lovey , 11 Years ago

Grade 12

2 Answers

Jitender Singh

Ans:
Hello Student,
Please find answer to your question below

$I = \int [\sqrt{tanx} + \sqrt{cotx}]dx$
$I = \int [\sqrt{\frac{sinx}{cosx}} + \sqrt{\frac{cosx}{sinx}}]dx$
$I = \int \frac{sinx + cosx}{\sqrt{sinx.cosx}}dx$
$sinx - cosx = t$
$(cosx + sinx)dx = dt$
$(sinx -cosx)^{2} = t^{2}$
$1 - 2sinx.cosx = t^{2}$
$sinx.cosx = \frac{1-t^{2}}{2}$
$I = \int \frac{\sqrt{2}}{\sqrt{1-t^{2}}}dt$
$I = \sqrt{2}sin^{-1}t + c$
$I = \sqrt{2}sin^{-1}(sinx -cosx) + c$

Approved

Last Activity: 11 Years ago

Yash Chourasiya

Dear Student

Please see the solution in the attachment.

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

Last Activity: 5 Years ago

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