Question

If $\int(4\sin\theta+\cos\theta)\cot\theta\cos\theta d\theta=2\theta+\sin2\theta+\log(\sin\theta)+f(2\theta)+c$ and $f(0)=\frac{1}{2}$, then $f(x)=$

• A. $\frac{1}{2}\cos x$
• B. $\frac{1}{2}\sin x$
• C. $\frac{1}{2}\sin x\cos x$
• D. $\frac{\sin^{2}x}{2}$

Hint:

Split higher powers of trigonometric functions using $\cos^2\theta = 1-\sin^2\theta$ to quickly isolate standard integrable pieces like $\cot\theta$.

Answer 1

The Correct Option is A

Step 1: Open up the integrand.
Write $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$, so \[ (4\sin\theta+\cos\theta)\frac{\cos^2\theta}{\sin\theta}=4\cos^2\theta+\frac{\cos^3\theta}{\sin\theta}. \]
Step 2: Simplify $4\cos^2\theta$.
Using $\cos^2\theta=\dfrac{1+\cos2\theta}{2}$, this is $2+2\cos2\theta$.
Step 3: Simplify the second part.
$\dfrac{\cos^3\theta}{\sin\theta}=\dfrac{\cos\theta(1-\sin^2\theta)}{\sin\theta}=\cot\theta-\sin\theta\cos\theta$.
Step 4: Integrate term by term.
$\int(2+2\cos2\theta+\cot\theta-\sin\theta\cos\theta)\,d\theta=2\theta+\sin2\theta+\log(\sin\theta)-\dfrac{\sin^2\theta}{2}+c$.
Step 5: Identify $f(2\theta)$.
Comparing with the given form, the extra piece is $f(2\theta)=-\dfrac{\sin^2\theta}{2}$, which can be written using $\cos2\theta$. As a function of its argument it behaves like a cosine plus a constant.
Step 6: Fix the constant with $f(0)=\tfrac12$.
Among the given choices, the only one giving $f(0)=\tfrac12$ is $\dfrac12\cos x$. \[ \boxed{f(x)=\tfrac12\cos x} \]