91Ó°ÊÓ

We need to evaluate the integral $$\int_{0}^{\pi / 4} \sec ^{4} \theta d \theta$$

Let's perform the substitution $$u = \sec \theta$$ and find the corresponding differential $$du = \sec \theta \tan \theta d\theta$$ We also need to change the limits of integration accordingly. When \(\theta=0\), we have \(u=\sec(0)=1\). When \(\theta=\frac{\pi}{4}\), we have \(u=\sec\left(\frac{\pi}{4}\right)=\sqrt{2}\). Thus, our integral now becomes: $$\int_{1}^{\sqrt{2}} u^3 \frac{du}{\sec^2\theta \tan\theta}$$

We can rewrite the integral in terms of \(u\) using the fact that \(\sec^2\theta=u^2\) and \(\tan^2\theta=\sec^2\theta-1\). Thus, we can substitute these expressions: $$\int_{1}^{\sqrt{2}} u^3 \frac{du}{u^2(u^2-1)}$$ Now, we can simplify the integral: $$\int_{1}^{\sqrt{2}} \frac{u}{(u^2-1)} du$$

To evaluate this integral, we will use partial fractions. We first write the expression \(\frac{u}{u^2-1}\) as the sum of two simpler fractions: $$\frac{u}{u^2-1}=\frac{A}{u-1}+\frac{B}{u+1}$$ Multiplying both sides by \((u^2-1)\), we get: $$u=A(u+1)+B(u-1)$$ Now, we will find the values of \(A\) and \(B\). Suppose \(u=1\), then the equation becomes: $$1=2A$$ which gives \(A=\frac{1}{2}\). Suppose \(u=-1\), then the equation becomes: $$-1=-2B$$ which gives \(B=\frac{1}{2}\). Now, substitute the values of A and B back into the expression: $$\frac{u}{u^2-1}=\frac{1}{2}\left[\frac{1}{u-1}+\frac{1}{u+1}\right]$$

Now, the integral becomes: $$\int_{1}^{\sqrt{2}}\frac{1}{2}\left[\frac{1}{u-1}+\frac{1}{u+1}\right]du$$ This integral can be split into two simpler integrals: $$\frac{1}{2}\left[\int_{1}^{\sqrt{2}}\frac{1}{u-1}du + \int_{1}^{\sqrt{2}}\frac{1}{u+1}du\right]$$ Now we can evaluate each integral separately: $$\frac{1}{2}\left[\ln|(u-1)|_{1}^{\sqrt{2}} + \ln|(u+1)|_{1}^{\sqrt{2}}\right]$$ Substitute the limits of integration: $$\frac{1}{2}\left[\ln\left|\frac{\sqrt{2}-1}{0}\right| + \ln\left|\frac{1+\sqrt{2}}{2}\right|\right]$$ Simplifying this expression, we find the value of the original integral: $$\frac{1}{2}\left[\ln(\sqrt{2}-1) - \ln(1) + \ln(1+\sqrt{2}) - \ln(2)\right]$$ Combine the terms: $$\frac{1}{2}\left[\ln\left(\frac{(\sqrt{2}-1)(1+\sqrt{2})}{2}\right)\right]$$ After simplification, the final answer is: $$\frac{1}{2}\ln\left(\frac{1}{2}\right)$$