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  • Evaluate the definite integrals in Exercises 1 to 20. 4 sin 2x from 0 to pi/4

Evaluate the definite integrals in Exercises 1 to 20.

    Q4.    \int_0^\frac{\pi}{4}\sin 2x dx

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Given integral: \int_0^\frac{\pi}{4}\sin 2x dx

Consider the integral \int \sin 2x dx

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

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

Now, by Second fundamental theorem of calculus, we have

I = f(\frac{\pi}{4})-f(0)

= \frac{-\cos 2(\frac{\pi}{4})}{2} + \frac{\cos 0}{2}

=\frac{1}{2} - 0

= \frac{1}{2}

Posted by

Divya Prakash Singh

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