`int_(0)^(pi//2) sin 2x tan^(-1) (sinx)dx = `

Options

A. `pi/2-1`
B. `1-pi/2`
C. `pi/4-1`
D. `1-pi/4`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \tan^{-1}(\sin x) \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
Using the double angle identity, we can rewrite \( \sin(2x) \) as \( 2 \sin(x) \cos(x) \):
\[ I = \int_{0}^{\frac{\pi}{2}} 2 \sin(x) \cos(x) \tan^{-1}(\sin x) \, dx \]

### Step 2: Change the Order of Multiplication
We can rearrange the terms in the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} 2 \sin(x) \tan^{-1}(\sin x) \cos(x) \, dx \]

### Step 3: Substitution
Let \( t = \sin(x) \). Then, \( dt = \cos(x) \, dx \). The limits change from \( x = 0 \) to \( x = \frac{\pi}{2} \) which corresponds to \( t = 0 \) to \( t = 1 \):
\[ I = \int_{0}^{1} 2t \tan^{-1}(t) \, dt \]

### Step 4: Integration by Parts
Now we will use integration by parts. Let:
- \( u = \tan^{-1}(t) \) so that \( du = \frac{1}{1+t^2} \, dt \)
- \( dv = 2t \, dt \) so that \( v = t^2 \)

Using integration by parts, we have:
\[ I = \left[ t^2 \tan^{-1}(t) \right]_{0}^{1} - \int_{0}^{1} t^2 \cdot \frac{1}{1+t^2} \, dt \]

### Step 5: Evaluate the Boundary Term
Evaluating the boundary term:
\[ \left[ t^2 \tan^{-1}(t) \right]_{0}^{1} = 1^2 \tan^{-1}(1) - 0^2 \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Step 6: Simplify the Remaining Integral
Now we need to evaluate:
\[ \int_{0}^{1} \frac{t^2}{1+t^2} \, dt \]
This can be simplified as:
\[ \int_{0}^{1} \left( 1 - \frac{1}{1+t^2} \right) dt = \int_{0}^{1} dt - \int_{0}^{1} \frac{1}{1+t^2} dt \]
Calculating these integrals:
\[ \int_{0}^{1} dt = 1 \]
\[ \int_{0}^{1} \frac{1}{1+t^2} dt = \tan^{-1}(t) \bigg|_{0}^{1} = \frac{\pi}{4} \]

### Step 7: Combine Results
Thus,
\[ \int_{0}^{1} \frac{t^2}{1+t^2} \, dt = 1 - \frac{\pi}{4} \]
Putting it all together:
\[ I = \frac{\pi}{4} - \left( 1 - \frac{\pi}{4} \right) = \frac{\pi}{4} - 1 + \frac{\pi}{4} = \frac{\pi}{2} - 1 \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \sin(2x) \tan^{-1}(\sin x) \, dx = \frac{\pi}{2} - 1 \]