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  • Evaluate the definite integrals in Exercises 1 to 20. 5 0 to pi/2 cos 2x

Evaluate the definite integrals in Exercises 1 to 20.

    Q5.    \int_0^\frac{\pi}{2}\cos 2x dx

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

Consider the integral \int \cos 2x dx

\int \cos 2x dx = \frac{\sin 2x }{2}

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

Now, by Second fundamental theorem of calculus, we have

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

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

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

Posted by

Divya Prakash Singh

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