Question:
The integral \( \int \frac{\cos x - \sin x}{10 + \sin 2x} dx \) is equal to:
The integral \( \int \frac{\cos x - \sin x}{10 + \sin 2x} dx \) is equal to:
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Always check if the numerator is a derivative of a part of the denominator or can be expressed as such by completing the square or using trigonometric identities.
Updated On: Jun 9, 2026
- \( \frac{1}{2} \log(10 + \sin 2x) + c \)
- \( \frac{1}{3} \log(10 + \sin 2x) + c \)
- \( \frac{1}{3} \tan^{-1} \left( \frac{\sin x + \cos x}{3} \right) + c \)
- \( \frac{1}{3} \tan^{-1} (10 + \sin 2x) + c \)
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The Correct Option is C
Solution and Explanation
Concept:
The numerator is the derivative of \( \sin x + \cos x \). We transform the denominator to express \( \sin 2x \) in terms of \( (\sin x + \cos x)^2 \).
Step 1: Manipulate the denominator.
Note that \( (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x \). Therefore, \( \sin 2x = (\sin x + \cos x)^2 - 1 \). Substituting this into the integral: $$I = \int \frac{\cos x - \sin x}{10 + (\sin x + \cos x)^2 - 1} dx = \int \frac{\cos x - \sin x}{9 + (\sin x + \cos x)^2} dx$$
Step 2: Perform u-substitution.
Let \( u = \sin x + \cos x \). Then, \( du = (\cos x - \sin x) dx \). $$I = \int \frac{1}{9 + u^2} du$$
Step 3: Integrate using the standard formula.
Using the formula \( \int \frac{1}{a^2 + u^2} du = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + c \): $$I = \frac{1}{3} \tan^{-1} \left( \frac{u}{3} \right) + c$$
Step 4: Substitute back.
$$I = \frac{1}{3} \tan^{-1} \left( \frac{\sin x + \cos x}{3} \right) + c$$ $$\boxed{\frac{1}{3} \tan^{-1} \left( \frac{\sin x + \cos x}{3} \right) + c}$$
Step 1: Manipulate the denominator.
Note that \( (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x \). Therefore, \( \sin 2x = (\sin x + \cos x)^2 - 1 \). Substituting this into the integral: $$I = \int \frac{\cos x - \sin x}{10 + (\sin x + \cos x)^2 - 1} dx = \int \frac{\cos x - \sin x}{9 + (\sin x + \cos x)^2} dx$$
Step 2: Perform u-substitution.
Let \( u = \sin x + \cos x \). Then, \( du = (\cos x - \sin x) dx \). $$I = \int \frac{1}{9 + u^2} du$$
Step 3: Integrate using the standard formula.
Using the formula \( \int \frac{1}{a^2 + u^2} du = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + c \): $$I = \frac{1}{3} \tan^{-1} \left( \frac{u}{3} \right) + c$$
Step 4: Substitute back.
$$I = \frac{1}{3} \tan^{-1} \left( \frac{\sin x + \cos x}{3} \right) + c$$ $$\boxed{\frac{1}{3} \tan^{-1} \left( \frac{\sin x + \cos x}{3} \right) + c}$$
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