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  • Evaluate the following: int 0 to pi/2 dx by a^2cos^2x plus b^2sin^2x (Hint: Divide Numerator and Denominator by cos4x)

Evaluate the following:

\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}
(Hint: Divide Numerator and Denominator by cos^4x)

 

Answers (1)

Given:\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}

Dividing Numerator and Denominator by cos^4x

\\ =\int_{0}^{\frac{\pi}{2}} \frac{\left(1 / \cos ^{4} x\right) d x}{\left(\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right) / \cos ^{2} x\right)^{2}} \\ \Rightarrow \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{4} x d x}{\left(a^{2}+b^{2} \tan ^{2} x\right)^{2}} \\ \Rightarrow \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x \sec ^{2} x d x}{\left(a^{2}+b^{2} \tan ^{2} x\right)^{2}}=\int_{0}^{\frac{\pi}{2}} \frac{\left(1+\tan ^{2} x\right) \sec ^{2} x d x}{\left(a^{2}+b^{2} \tan ^{2} x\right)^{2}} \\ \Rightarrow \text { Put } \tan x=t

\\ \Rightarrow \sec ^{2} x d x=d t \\ \Rightarrow A t x=0=>t=0 \text { and at } x=\frac{\pi}{2}=>t=\infty \\ \Rightarrow \int_{0}^{\frac{\pi}{2}} \frac{\left(1+\tan ^{2} x\right) \sec ^{2} x d x}{\left(a^{2}+b^{2} \tan ^{2} x\right)^{2}}=\int_{0}^{\infty} \frac{\left(1+t^{2}\right) d t}{\left(a^{2}+b^{2} t^{2}\right)^{2}} \\ \Rightarrow \int_{0}^{\infty} \frac{\left(1+t^{2}\right) d t}{\left(a^{2}+b^{2} t^{2}\right)^{2}}=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(b^{2}+b^{2} t^{2}\right) d t}{\left(a^{2}+b^{2} t^{2}\right)^{2}}

\\ =\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(b^{2}+b^{2} t^{2}\right) d t}{\left(a^{2}+b^{2} t^{2}\right)^{2}}=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(a^{2}+b^{2} t^{2}\right)+\left(b^{2}-a^{2}\right)}{\left(a^{2}+b^{2} t^{2}\right)^{2}} d t \\ \Rightarrow \frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(a^{2}+b^{2} t^{2}\right)+\left(b^{2}-a^{2}\right)}{\left(a^{2}+b^{2} t^{2}\right)^{2}} d t=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{\left(a^{2}+b^{2} t^{2}\right)} d t+\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(b^{2}-a^{2}\right)}{\left(a^{2}+b^{2} t^{2}\right)^{2}} d t \\ \Rightarrow \text { Let } I=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{\left(a^{2}+b^{2} t^{2}\right)} d t+\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(b^{2}-a^{2}\right)}{\left(a^{2}+b^{2} t^{2}\right)^{2}} d t \ldots(1)

\\ =\operatorname{Let} I_{1}=\int_{0}^{\infty} \frac{1}{\left(a^{2}+b^{2} t^{2}\right)} d t \\ =\int_{0}^{\infty} \frac{1}{\left(a^{2}+b^{2} t^{2}\right)} d t=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{\left(\left(a^{2} / b^{2}\right)+t^{2}\right)} d t \\ =1_{1}=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{\left(\left(a^{2} / b^{2}\right)+t^{2}\right)} d t=\frac{1}{b^{2}}\left(\frac{b}{a}\right)\left[\tan ^{-1}\left(\frac{b t}{a}\right)\right]_{0}^{\infty}=\frac{\pi}{2 a b} \\ \Rightarrow I_{1}=\frac{\pi}{2 a b} \ldots .(2)

 

\\ \text { Let } I_{2}=\int_{0}^{\infty} \frac{1}{\left(a^{2}+b^{2} t^{2}\right)^{2}} d t \\ \Rightarrow \text { let } b t=a \tan \theta \\ \Rightarrow b d t=\operatorname{asec}^{2} \theta d \theta \\ =I_{2}=\frac{1}{b} \int_{0}^{\frac{\pi}{2}} \frac{\operatorname{asec}^{2} \theta d \theta}{\left(a^{2}+a^{2} \tan ^{2} \theta\right)^{2}}=\frac{1}{b} \int_{0}^{\frac{\pi}{2}} \frac{a \sec ^{2} \theta d \theta}{a^{4}\left(1+\tan ^{2} \theta\right)^{2}}=\frac{1}{a^{3} b} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta d \theta}{\sec ^{4} \theta}

\\ =\frac{1}{a^{3} b} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta \mathrm{d} \theta}{\sec ^{4} \theta}=\frac{1}{a^{3} \mathrm{~b}} \int_{0}^{\frac{\pi}{2}} \cos ^{2} \theta \mathrm{d} \theta=\frac{1}{2 a^{3} \mathrm{~b}} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 \theta) \mathrm{d} \theta \\ \Rightarrow \frac{1}{2 a^{3} b} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 \theta) \mathrm{d} \theta=\frac{1}{2 a^{3} b}\left[\theta+\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}}=\frac{\pi}{4 a^{3} b} \ldots(3) \\ \Rightarrow I=\frac{1}{b^{2}}\left(I_{1}+\left(b^{2}-a^{2}\right) \mathrm{I}_{2}\right)=\frac{1}{b^{2}}\left(\frac{\pi}{2 a b}+\left(b^{2}-a^{2}\right) \frac{\pi}{4 a^{3} b}\right)

\\\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}=\frac{\pi}{2 a b^{3}}\left(1+\left(\frac{b^{2}}{a^{2}}-1\right) \pi\right)

 

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