Question

$\int_{0}^{\pi/4} \frac{\sec x}{3\cos x + 4\sin x} dx =$

• A. $\log(7/3)$
• B. $\frac{1}{4}\log(7/3)$
• C. $\frac{1}{4}\log 7$
• D. $\log 7$

Hint:

For integrals involving trigonometric functions, a common strategy is to divide the numerator and denominator by a power of $\cos x$ or $\sin x$ to convert the expression into terms of $\tan x$ and $\sec^2 x$, which sets up a simple u-substitution.

Answer 1

The Correct Option is B

Let the integral be $I$.
$I = \int_{0}^{\pi/4} \frac{\sec x}{3\cos x + 4\sin x} dx$.
Multiply the numerator and the denominator by $\sec x$:
$I = \int_{0}^{\pi/4} \frac{\sec^2 x}{(3\cos x + 4\sin x)\sec x} dx = \int_{0}^{\pi/4} \frac{\sec^2 x}{3 + 4\frac{\sin x}{\cos x}} dx$.
$I = \int_{0}^{\pi/4} \frac{\sec^2 x}{3 + 4\tan x} dx$.
Now, we use a substitution. Let $u = 3 + 4\tan x$.
Then, $du = 4\sec^2 x dx$, which means $\sec^2 x dx = \frac{du}{4}$.
We also need to change the limits of integration:
When $x=0$, $u = 3 + 4\tan(0) = 3+0 = 3$.
When $x=\pi/4$, $u = 3 + 4\tan(\pi/4) = 3+4(1) = 7$.
The integral in terms of $u$ becomes:
$I = \int_{3}^{7} \frac{1}{u} \left(\frac{du}{4}\right) = \frac{1}{4} \int_{3}^{7} \frac{1}{u} du$.
$I = \frac{1}{4} [\log|u|]_{3}^{7}$.
$I = \frac{1}{4} (\log 7 - \log 3)$.
Using the properties of logarithms, we get:
$I = \frac{1}{4} \log\left(\frac{7}{3}\right)$.