Question

Use the approaches discussed in this section to evaluate the following integrals. $$\int_{0}^{\pi / 4} 3 \sqrt{1+\sin 2 x} d x$$

Short answer

Question: Evaluate the following integral: $$\int_{0}^{\pi / 4} 3 \sqrt{1+\sin 2 x} d x$$ Answer: $$\int_{0}^{\pi / 4} 3 \sqrt{1+\sin 2 x} d x = \frac{\sqrt{2}}{2} + 2\sqrt{2}\tan^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)$$

Step-by-step solution

Simplify the integrand using trigonometric identity

Recall the double-angle formula for sine: \(\sin{2x} = 2\sin{x}\cos{x}\). Using this, we can rewrite the integrand as: $$3\sqrt{1 + 2\sin{x}\cos{x}}$$

Perform substitution

We can make the following substitution: $$u = \sin{x}$$ $$du = \cos{x}dx$$ Now replace x with the corresponding bounds for u: When \(x = 0\), \(u = \sin{0} = 0\) When \(x = \frac{\pi}{4}\), \(u = \sin{\frac{\pi}{4}} = \frac{\sqrt2}{2}\) Now, rewrite the integral using the substitution: $$\int_0^{\frac{\sqrt{2}}{2}} 3\sqrt{1+ 2u(\cos{x})} du$$

Eliminate remaining x term

Now we need to eliminate the \(\cos{x}\) term from the integral. We can use the Pythagorean identity \(\sin^2{x} + \cos^2{x} = 1\) and in terms of u, this becomes: $$u^2+\cos^2{x}=1$$ $$\cos{x}=\pm\sqrt{1-u^2}$$ Here, the value of x lies in the first quadrant (\(0\leq x\leq \frac{\pi}{4}\)), so the cosine is positive, thus we have: $$\cos{x} = \sqrt{1-u^2}$$ Substitute this back into the integral: $$\int_0^{\frac{\sqrt{2}}{2}} 3\sqrt{1+ 2u\sqrt{1-u^2}} du$$

Perform integration

The antiderivative of the integrand can be calculated as: $$3\int_0^{\frac{\sqrt{2}}{2}} \sqrt{1+ 2u\sqrt{1-u^2}} du = 2\sqrt{2}\left([u\sqrt{1-u^2}+\sqrt{2}\tan^{-1} \left(\frac{u}{\sqrt{2-u^2}}\right)]_0^{\frac{\sqrt{2}}{2}}\right)$$

Evaluate the antiderivative

Evaluate the antiderivative at the bounds: $$2\sqrt{2}\left([\frac{\sqrt{2}}{2}\sqrt{1-\frac{1}{2}}+\sqrt{2}\tan^{-1} \left(\frac{\frac{\sqrt{2}}{2}}{\sqrt{2-\frac{1}{2}}}\right)]-[0\sqrt{1-0}+\sqrt{2}\tan^{-1} \left(\frac{0}{\sqrt{2-0}}\right)]\right)$$ $$= 2\sqrt{2}\left([\frac{\sqrt{2}}{2}\sqrt{\frac{1}{2}}+\sqrt{2}\tan^{-1} \left(\frac{\sqrt{2}}{\sqrt{3}}\right)]\right)$$ $$= 2\sqrt{2}\left(\frac{\sqrt{2}}{4}+\sqrt{2}\tan^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)\right)$$ The solution to the integral is: $$\int_{0}^{\pi / 4} 3 \sqrt{1+\sin 2 x} d x = \frac{\sqrt{2}}{2} + 2\sqrt{2}\tan^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)$$