Question

π/2∫ sin²x/sin x+cosx dx =1/√2 log(√2+1) ,Prove it.0

Answer 1

Answer:

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Answer 2

Answer:

Proved that,  $\int\limits^\frac{\pi }{2} _0 {\frac{Sin^{2}x }{Sinx+Cosx} } \, dx=\frac{1}{\sqrt{2} } ln(\sqrt{2}+1 )$

Step-by-step explanation:

We have to prove that, $\int\limits^\frac{\pi }{2} _0 {\frac{Sin^{2}x }{Sinx+Cosx} } \, dx=\frac{1}{\sqrt{2} } ln(\sqrt{2}+1 )$

Let us assume, I = $\int\limits^\frac{\pi }{2} _0 {\frac{Sin^{2}x }{Sinx+Cosx} } \, dx$ ...... (1)

= $\int\limits^\frac{\pi }{2} _0 {\frac{Sin^{2}(\frac{\pi }{2}- x) }{Sin(\frac{\pi }{2} -x)+Cos(\frac{\pi }{2} -x)} } \, dx$ {Using properties of definite integral}

⇒ I = $\int\limits^\frac{\pi }{2} _0 {\frac{Cos^{2}x }{Cosx+Sinx} } \, dx$ .......(2)

Now, by (1) + (2) we get

2I = $\int\limits^\frac{\pi }{2} _0 {\frac{Sin^{2}x }{Sinx+Cosx} } \, dx +\int\limits^\frac{\pi }{2} _0 {\frac{Cos^{2}x }{Sinx+Cosx} } \, dx$

⇒ 2I = $\int\limits^\frac{\pi }{2} _0 {\frac{1}{Sinx+Cosx} } \, dx$

⇒ I = $\frac{1}{2} \int\limits^\frac{\pi }{2} _0 {\frac{1}{Sinx+Cosx} } \, dx$........ (3)

Now we will do the indefinite integral first, then will put the limits to get the definite value.

Let, I' =∫$\frac{dx}{Sinx+Cosx}$

=∫$\frac{dx}{2Sin\frac{x}{2}Cos\frac{x}{2}+2Cos^{2}\frac{x}{2}-1    }$

= ∫$\frac{Sec^{2}\frac{x}{2}dx  }{2tan\frac{x}{2}+2-Sec^{2}\frac{x}{2}   }$

= ∫$\frac{2d[tan\frac{x}{2} ]}{2tan\frac{x}{2}+1-tan^{2}\frac{x}{2}   }$

= -2∫$\frac{d[tan\frac{x}{2} ]}{tan^{2}\frac{x}{2}-2tan\frac{x}{2}-1   }$

= -2∫$\frac{dz}{z^{2}-2z-1 }$ {Where, z= tan$\frac{x}{2}$}

= -2∫$\frac{d(z-1)}{(z-1)^{2}-(\sqrt{2}) ^{2}  }$

[Using the formula, ∫$\frac{dx}{x^{2}-a^{2}  }=\frac{1}{2a}ln\frac{x-a}{x+a}$]

= $\frac{-2}{2\sqrt{2} }ln\frac{z-1-\sqrt{2} }{z-1+\sqrt{2} }$ +C {Where C is an integration constant}

= $\frac{-1}{\sqrt{2} } ln\frac{tan\frac{x}{2}-1-\sqrt{2}  }{tan\frac{x}{2}-1+\sqrt{2}  }+C$

Now, putting the limits we will get the value of I from equation (3).

Hence, I = $\frac{1}{2} [\frac{-1}{\sqrt{2} } ln\frac{tan\frac{x}{2}-1-\sqrt{2}  }{tan\frac{x}{2}-1+\sqrt{2}  }]_{0} ^{\frac{\pi }{2} }$ { C will be cancelled}

= $\frac{-1}{2\sqrt{2} } [0-ln\frac{\sqrt{2}+1 }{\sqrt{2}-1 } ]$

= $\frac{1}{2\sqrt{2} }ln\frac{(\sqrt{2}+1 )^{2} }{1}$

= $\frac{1}{\sqrt{2} } ln(\sqrt{2}+1 )$

Hence, proved.