Question

Evaluate the Equation \(\int\limits_0^{\pi /2} {\frac{{\cos t}}{{\sqrt {1 + {{\sin }^2}t} }}dt} \)

Short answer

The value of the integral is ln (1+√2)

Suppose sint = tanθ. Differentiate it and convert the given integral with respect to θ. Integrate it and then evaluate the value of the integral.

Step-by-step solution

Given Data

Put sint = tanθ and convert the integral in θ

Let sint = tanθ

Costdt = sec²θ dθ

When t = 0, θ = tan^-1(sin0) = 0

When t = π/2, θ = tan^-1 (sin π/2) = π/4

\(\begin{aligned}{l}\int\limits_0^{\pi /2} {\frac{{\cos t}}{{\sqrt {1 + {{\sin }^2}t} }}dt} &= \int\limits_0^{\pi /4} {\frac{{{{\sec }^2}\theta }}{{\sqrt {1 + {{\tan }^2}\theta } }}d\theta } &\\\int\limits_0^{\pi /4} {\frac{{{{\sec }^2}\theta }}{{\sec \theta }}d\theta } &\\\int\limits_0^{\pi /4} {\sec \theta d\theta } \end{aligned}\)(Since 1+tan² θ = sec² θ)

Evaluation

Evaluate the value of the integral

\(\begin{aligned}&{l}\int\limits_0^{\pi /4} {\sec \theta d\theta = (\ln \left| {\sec \theta + \tan \theta } \right|} )_0^{\pi /4}\\ &= \ln \left| {\sec \pi /4 + \tan \pi /4} \right| - \ln \left| {\sec 0 + \tan 0} \right|\\ &= \ln \left| {\sqrt 2 + 1} \right| - \ln \left| {1 + 0} \right|\\ &= \ln (1 + \sqrt 2 )\\\end{aligned}\)

Hence, \(\int\limits_0^{\pi /2} {\frac{{\cos t}}{{1 + {{\sin }^2}t}}dt = \ln (1 + \sqrt 2 )} \)