The value of the integral Ι=∫0∞dx/(1+x^20201+x^2) is equal to kπ, then the value of 16k is equal to

Question

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Correct answer: 4. Put tan^(-1)x=θ ⇒x=tan⁡θ ⇒dx=sec^2θdθ ⇒Ι=∫0π/2(sec^2θ)/(1+tan^2020θsec^2θ)dθ =∫0(π)/2(dθ)/(1+tan^2020θ) Ι=∫0(π)/2(cos^2020θ)/(cos^2020θ+sin^2020θ)dθ Applying a+b-x property and adding, we get, 2Ι=∫0(π)/2(cos^2020θ+sin^2020θ)/(cos^2020θ+sin^2020θ)dθ ⇒2Ι=θ0(π)/2=(π)/2-0 ⇒Ι=(π)/4 ⇒k=1/4⇒16k=4

The value of the integral $Ι = \int_{0}^{\infty} \frac{d x}{(1 + x^{2020}) (1 + x^{2})}$ is equal to $k π$ , then the value of $16 k$ is equal to

Solution

More Definite Integration Questions

The value of the integral Ι=∫0∞dx1+x20201+x2 is equal to kπ, then the value of 16k is equal to

Solution

More Definite Integration Questions

The value of the integral $Ι = \int_{0}^{\infty} \frac{d x}{(1 + x^{2020}) (1 + x^{2})}$ is equal to $k π$ , then the value of $16 k$ is equal to