Math, asked by Anonymous, 1 year ago

Evaluate!!

Definite integrals!!!


Please help.

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Answered by hukam0685
3
hello

please find the solution in attachment
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Anonymous: Omg Guru Ji.
Anonymous: BTW thank you so much for your answer Guru Ji.
hukam0685: yes ,i told you they are very long to answer
Anonymous: Yes Guru Ji. But i am so much thankful to you for your help. :heart:
Answered by rohitkumargupta
7

HELLO DEAR,



\bold{I = \int\limits^{\frac{\pi}{4}}_0 {(\sqrt{tanx} + \sqrt{cotx})} \, dx }



\bold{I = \int\limits^{\frac{\pi}{4}}_0 {(\frac{\sqrt{sinx}}{\sqrt{cosx}} + \frac{\sqrt{cosx}}{\sqrt{sinx}})}\,dx}



\bold{I = \int\limits^{\frac{\pi}{4}}_0 {(\frac{sinx + cosx}{\sqrt{sinx*cosx}})}\,dx}



\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{4}}_0 {\frac{sinx + cosx}{\sqrt{2sinxcosx}}}\,dx}



\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{4}}_0 {\frac{sinx + cosx}{\sqrt{sin2x}}}\,dx}



\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{4}}_0 {\frac{sinx + cosx}{\sqrt{1 - (1 - sin2x)}}}\,dx}



\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{4}}_0 {\frac{sinx + cosx}{\sqrt{1 - (sinx - cosx)^2}}}\,dx}



let t = (sinx - cosx)


{\Rightarrow } dt = (cosx + sinx).dx



Also, [x = 0 {\Rightarrow } t = -1] and [x = π/4 {\Rightarrow }t = 0 ]



\bold{\therefore, I = \sqrt{2}\int\limits^{0}_{-1} {\frac{dt}{\sqrt{1 - t^2}}}}



\bold{\Rightarrow I = \sqrt{2}[sin^{-1}t]^{0}_{-1}}



\bold{\Rightarrow I = \sqrt{2}[sin^{-1}(0) - sin^{-1}(-1)]}



\bold{\Rightarrow I = \sqrt{2}[0 - (-\pi/2)]}



\bold{\Rightarrow I = \sqrt{2}*\frac{\pi}{2}}



\bold{\Rightarrow I = \frac{\pi}{\sqrt{2}}}



I HOPE ITS HELP YOU DEAR,


THANKS

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