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  • Evaluate the definite integrals in Exercises 1 to 20. 12

Evaluate the definite integrals in Exercises 1 to 20.

    Q12.    \int_0^\frac{\pi}{2}\cos^2 x dx

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Given integral: \int_0^\frac{\pi}{2}\cos^2 x dx

Consider the integral \int \cos^2 x dx

\int \cos^2 x dx = \int \frac{1+\cos 2x}{2} dx = \frac{x}{2}+\frac{\sin 2x }{4}

= \frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )

So, we have the function of x, f(x) =\frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )

Now, by Second fundamental theorem of calculus, we have

I = f(\frac{\pi}{2}) -f(0)

= \frac{1}{2}\left \{ \left ( \frac{\pi}{2}-\frac{\sin \pi}{2} \right ) -\left ( 0+\frac{\sin 0}{2} \right ) \right \}

= \frac{1}{2}\left \{ \frac{\pi}{2}+0-0-0 \right \}

= \frac{\pi}{4}

Posted by

Divya Prakash Singh

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