Question

1. $int_{0}^{pi/4}\frac{2 + cos^{3}u}{cos^{2}u}du=$ video example: solving a similar problem submit answer get help practice similar

Explanation:

Step1: Split the integral

We split $\int_{0}^{\pi/4}\frac{2 + \cos^{3}u}{\cos^{2}u}du$ into two - integrals: $\int_{0}^{\pi/4}\frac{2}{\cos^{2}u}du+\int_{0}^{\pi/4}\frac{\cos^{3}u}{\cos^{2}u}du=\int_{0}^{\pi/4}2\sec^{2}u\ du+\int_{0}^{\pi/4}\cos u\ du$.

Step2: Integrate each part

For $\int_{0}^{\pi/4}2\sec^{2}u\ du$, since $\int\sec^{2}u\ du=\tan u + C$, then $\int_{0}^{\pi/4}2\sec^{2}u\ du=2[\tan u]_{0}^{\pi/4}=2(\tan\frac{\pi}{4}-\tan0)=2(1 - 0)=2$.
For $\int_{0}^{\pi/4}\cos u\ du$, since $\int\cos u\ du=\sin u + C$, then $\int_{0}^{\pi/4}\cos u\ du=[\sin u]_{0}^{\pi/4}=\sin\frac{\pi}{4}-\sin0=\frac{\sqrt{2}}{2}-0=\frac{\sqrt{2}}{2}$.

Step3: Combine the results

Add the results of the two - integrals: $2+\frac{\sqrt{2}}{2}=\frac{4 + \sqrt{2}}{2}$.

Answer:

$\frac{4+\sqrt{2}}{2}$