Question

\(\int\limits_{0}^{\frac{\pi}{2}}\frac{\cos x\sin x}{1+\sin x}\ dx\) is equal to

• A. log 2 -1
• B. -log 2
• C. log 2
• D. 1 -log 2

Answer 1

The Correct Option is D

We are given the integral: \[ \int\limits_{0}^{\frac{\pi}{2}}\frac{\cos x \sin x}{1+\sin x} \, dx \] 

Let us simplify the integrand:
Use substitution: Let \( u = 1 + \sin x \)
Then, \[ \frac{du}{dx} = \cos x \Rightarrow du = \cos x \, dx \] Also, \( \sin x = u - 1 \), so: \[ \cos x \, dx = du, \quad \sin x = u - 1 \Rightarrow \cos x \sin x \, dx = (u - 1) \, du \] Thus the integral becomes: \[ \int \frac{(u - 1)}{u} \, du = \int \left(1 - \frac{1}{u} \right) du = u - \ln|u| \] Now change limits: - When \( x = 0 \), \( u = 1 + \sin 0 = 1 \) - When \( x = \frac{\pi}{2} \), \( u = 1 + \sin \frac{\pi}{2} = 2 \) So, \[ \int\limits_{1}^{2} \left(1 - \frac{1}{u} \right) du = \left[ u - \ln|u| \right]_1^2 = (2 - \ln 2) - (1 - \ln 1) = 2 - \ln 2 - 1 = 1 - \ln 2 \] 

Final Answer: \(1 - \log 2\)

Answer 2

Let the integral be \(I\). 

\[ I = \int\limits_{0}^{\frac{\pi}{2}}\frac{\cos x\sin x}{1+\sin x}\ dx \]

We use the substitution method. Let \(u = 1 + \sin x\).

Then, differentiate with respect to \(x\):

\[ \frac{du}{dx} = \cos x \]

So, \(du = \cos x \, dx\).

We also need to express \(\sin x\) in terms of \(u\). From \(u = 1 + \sin x\), we get \(\sin x = u - 1\).

Now, we change the limits of integration:

When \(x = 0\), \(u = 1 + \sin(0) = 1 + 0 = 1\). The lower limit becomes \(u=1\).

When \(x = \frac{\pi}{2}\), \(u = 1 + \sin\left(\frac{\pi}{2}\right) = 1 + 1 = 2\). The upper limit becomes \(u=2\).

Substitute \(u\), \(du\), and \(\sin x\) into the integral:

\[ I = \int_{1}^{2} \frac{(u-1)}{u} \, du \]

Simplify the integrand:

\[ I = \int_{1}^{2} \left(\frac{u}{u} - \frac{1}{u}\right) \, du = \int_{1}^{2} \left(1 - \frac{1}{u}\right) \, du \]

Now, integrate with respect to \(u\):

\[ I = \left[ u - \ln|u| \right]_{1}^{2} \]

Evaluate the antiderivative at the upper and lower limits:

\[ I = (2 - \ln|2|) - (1 - \ln|1|) \]

Since the limits are positive, we can remove the absolute value signs. We know that \(\ln(1) = 0\).

\[ I = (2 - \ln(2)) - (1 - 0) \]

\[ I = 2 - \ln(2) - 1 \]

\[ I = \mathbf{1 - \ln(2)} \]

Assuming 'log' represents the natural logarithm ('ln'), the result is \(1 - \log 2\).

Comparing this with the given options, the correct option is:

1 - log 2