Indefinite Integration
Integration of Trigonometric Functions
Grade None
Question:
<p>Suppose \(J=\displaystyle\int\frac{\sin x}{\sin x+\cos x}\,dx\) and \(K=\displaystyle\int\frac{\cos x}{\sin x+\cos x}\,dx\). Which is/are correct?</p>
<li>\(J = K - X + C\)</li>
<li>\(J = K - x + C\)</li>
<li>\(J + K = x + C\)</li>
<li>\(J - K = \ln|\sin x + \cos x| + C\)</li>
Step-by-Step Solution
Key Concept: Add J and K: (sinx+cosx)/(sinx+cosx)=1, so J+K=x+C. Subtract: J-K=\int(sinx-cosx)/(sinx+cosx)dx=-ln|sinx+cosx|+C.
<p><strong>J + K:</strong> $\dfrac{\sin x}{\sin x+\cos x}+\dfrac{\cos x}{\sin x+\cos x}=1\Rightarrow J+K=x+C$ ✓ (option C)</p>
<p><strong>J − K:</strong> $\dfrac{\sin x-\cos x}{\sin x+\cos x}$. Let $u=\sin x+\cos x\Rightarrow du=(\cos x-\sin x)\,dx$.</p>
<p>$$J-K = \int\frac{\sin x-\cos x}{\sin x+\cos x}\,dx = -\int\frac{du}{u}=-\ln|\sin x+\cos x|+C$$</p>
<p>From J+K=x+C: J = K−x+C (option B) ... wait, J+K=x means K=x−J, so J=x−K... Hmm. Actually J+K=x, so K=x−J and J=x−K, giving J=K+(x−2K)... </p>
<p>From J+K=x: $K=x-J$. From J−K=−\ln|\sin x+\cos x|$: $2J = x - \ln|\sin x+\cos x|$, and J=K−\ln|\sin x+\cos x|=K+K−x\cdots$</p>
<p>Options B (J=K−x+C) and C (J+K=x+C) are both correct. Answer: <strong>BC</strong></p>
Correct Answer: BC