The value of int_0^{frac{pi}{2}} frac{dx}{1+t | vedclass

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(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{dx}{1+	an^3 x}$.We can rewrite the integrand as: $I = \int_0^{\frac{\pi}{2}} \frac{\cos^3 x}{\cos^3 x + \si

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{dx}{1+	an^3 x}$.We can rewrite the integrand as: $I = \int_0^{\frac{\pi}{2}} \frac{\cos^3 x}{\cos^3 x + \sin^3 x} dx$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have:$I = \int_0^{\frac{\pi}{2}} \frac{\cos^3(\frac{\pi}{2}-x)}{\cos^3(\frac{\pi}{2}-x) + \sin^3(\frac{\pi}{2}-x)} dx = \int_0^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} dx$.Adding the two expressions for $I$:$2I = \int_0^{\frac{\pi}{2}} \frac{\cos^3 x + \sin^3 x}{\cos^3 x + \sin^3 x} dx = \int_0^{\frac{\pi}{2}} 1 dx = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$.Therefore,$I = \frac{\pi}{4}$.

The value of $\int_0^{\frac{\pi}{2}} \frac{dx}{1+\tan^3 x}$ is:

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