Question:
The integral \( \int \frac{\cos^3 x}{(1+\sin x)^4} dx \) is equal to:
The integral \( \int \frac{\cos^3 x}{(1+\sin x)^4} dx \) is equal to:
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When the denominator contains \(1+\sin x\), try the substitution \(u=1+\sin x\). Also use the identity \(\cos^2x=1-\sin^2x\) to reduce higher powers of cosine.
Updated On: Jun 9, 2026
- \( -\frac{\cos^4 x}{5(1+\sin x)^5} + c \)
- \( \frac{\cos^4 x}{5(1+\sin x)^5} + c \)
- \( \frac{\cos^4 x}{4(1+\sin x)^4} + c \)
- \( -\frac{\cos^4 x}{4(1+\sin x)^4} + c \)
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The Correct Option is D
Solution and Explanation
Concept:
Use the substitution \(u=1+\sin x\). The derivative \(du=\cos x\,dx\) appears naturally in the integrand, allowing us to convert the trigonometric integral into a simple algebraic integral.
Step 1: Rewrite the numerator.
\[ I=\int \frac{\cos^3 x}{(1+\sin x)^4}\,dx =\int \frac{\cos^2x\cos x}{(1+\sin x)^4}\,dx \] Using \[ \cos^2x=1-\sin^2x, \] we get \[ I=\int \frac{(1-\sin^2x)\cos x}{(1+\sin x)^4}\,dx. \] Factor the numerator: \[ 1-\sin^2x=(1-\sin x)(1+\sin x). \] Hence, \[ I=\int \frac{(1-\sin x)\cos x}{(1+\sin x)^3}\,dx. \]
Step 2: Apply substitution.
Let \[ u=1+\sin x. \] Then \[ du=\cos x\,dx, \qquad 1-\sin x=2-u. \] Substituting, \[ I=\int \frac{2-u}{u^3}\,du. \] \[ I=\int \left(\frac{2}{u^3}-\frac{1}{u^2}\right)du. \]
Step 3: Integrate term-by-term.
\[ I = 2\int u^{-3}\,du - \int u^{-2}\,du. \] \[ I = 2\left(\frac{u^{-2}}{-2}\right) - \left(\frac{u^{-1}}{-1}\right) +c. \] \[ I = -\frac{1}{u^2} +\frac{1}{u} +c. \] Substituting \(u=1+\sin x\), \[ I = -\frac{1}{(1+\sin x)^2} +\frac{1}{1+\sin x} +c. \] \[ I = \frac{\sin x}{(1+\sin x)^2} +c. \]
Step 4: Express the result in the form of the options.
Using \[ \cos^2x=(1-\sin x)(1+\sin x), \] we have \[ \frac{\sin x}{(1+\sin x)^2} = -\frac{\cos^4x}{4(1+\sin x)^4} +\text{constant}. \] Therefore, the antiderivative can be written as \[ \boxed{ I= -\frac{\cos^4x}{4(1+\sin x)^4}+c } \] which matches option (D).
Step 1: Rewrite the numerator.
\[ I=\int \frac{\cos^3 x}{(1+\sin x)^4}\,dx =\int \frac{\cos^2x\cos x}{(1+\sin x)^4}\,dx \] Using \[ \cos^2x=1-\sin^2x, \] we get \[ I=\int \frac{(1-\sin^2x)\cos x}{(1+\sin x)^4}\,dx. \] Factor the numerator: \[ 1-\sin^2x=(1-\sin x)(1+\sin x). \] Hence, \[ I=\int \frac{(1-\sin x)\cos x}{(1+\sin x)^3}\,dx. \]
Step 2: Apply substitution.
Let \[ u=1+\sin x. \] Then \[ du=\cos x\,dx, \qquad 1-\sin x=2-u. \] Substituting, \[ I=\int \frac{2-u}{u^3}\,du. \] \[ I=\int \left(\frac{2}{u^3}-\frac{1}{u^2}\right)du. \]
Step 3: Integrate term-by-term.
\[ I = 2\int u^{-3}\,du - \int u^{-2}\,du. \] \[ I = 2\left(\frac{u^{-2}}{-2}\right) - \left(\frac{u^{-1}}{-1}\right) +c. \] \[ I = -\frac{1}{u^2} +\frac{1}{u} +c. \] Substituting \(u=1+\sin x\), \[ I = -\frac{1}{(1+\sin x)^2} +\frac{1}{1+\sin x} +c. \] \[ I = \frac{\sin x}{(1+\sin x)^2} +c. \]
Step 4: Express the result in the form of the options.
Using \[ \cos^2x=(1-\sin x)(1+\sin x), \] we have \[ \frac{\sin x}{(1+\sin x)^2} = -\frac{\cos^4x}{4(1+\sin x)^4} +\text{constant}. \] Therefore, the antiderivative can be written as \[ \boxed{ I= -\frac{\cos^4x}{4(1+\sin x)^4}+c } \] which matches option (D).
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