Question

Evaluate \[ \int\frac{\cos x}{\sqrt{16\cos^2x+9}}dx \]

• A. \(\frac14\sinh^{-1}\left(\frac{4\sin x}{5}\right)+c\)
• B. \(\frac14\sin^{-1}\left(\frac{4\sin x}{5}\right)+c\)
• C. \(\frac14\cosh^{-1}\left(\frac{4\sin x}{3}\right)+c\)
• D. \(\frac14\cos^{-1}\left(\frac{4\cos x}{3}\right)+c\)

Hint:

Whenever square root has quadratic structure, convert immediately to standard inverse trigonometric or inverse hyperbolic form.

Answer 1

The Correct Option is A

Concept: For expressions involving square root quadratic form \[ \sqrt{a^2+u^2} \] inverse hyperbolic substitution works naturally.

Step 1: Substitute variable.
Let \[ t=\sin x \] Then \[ dt=\cos xdx \] Integral becomes \[ I= \int \frac{dt} {\sqrt{16(1-t^2)+9}} \] \[ = \int \frac{dt} {\sqrt{25-16t^2}} \]

Step 2: Rewrite standard form.
Take \[ u=\frac{4t}{5} \] Then \[ dt=\frac54du \] Thus \[ I = \frac14 \int \frac{du}{\sqrt{1-u^2}} \] Standard formula gives \[ = \frac14 \sinh^{-1} \left( \frac{4t}{5} \right) +c \] Substitute back. \[ = \frac14 \sinh^{-1} \left( \frac{4\sin x}{5} \right) +c \] Hence \[ \boxed{ \frac14 \sinh^{-1} \left( \frac{4\sin x}{5} \right)+c } \]