Integrate ( x^2cosx)/(1+sinx)^2 dx
with limits form 0 ti pi/4
Solution plz
To integrate the given expression, we will use the substitution method. Let's go through the steps:
1. Start by letting u = 1 + sin(x). Therefore, du/dx = cos(x), and dx = du/cos(x).
2. Substitute these values in the integral:
∫(x^2cos(x))/(1+sin(x))^2 dx = ∫(x^2cos(x))/(u^2) * (du/cos(x))
= ∫(x^2/cos(x)) * (1/u^2) * du
3. Simplify the expression:
The numerator becomes x^2/cos(x) = x^2sec(x).
The denominator becomes 1/u^2 = 1/(1+sin(x))^2.
4. Rewrite the integral using the new variables:
∫(x^2cos(x))/(1+sin(x))^2 dx = ∫(x^2sec(x)) * (1/(1+sin(x))^2) du
5. Now, let's deal with the limits of integration:
When x = 0, u = 1 + sin(0) = 1 + 0 = 1.
When x = π/4, u = 1 + sin(π/4) = 1 + 1/√2 = (1+√2)/√2.
Therefore, the limits of integration will be from u = 1 to u = (1+√2)/√2.
6. Rewrite the integral with the new limits:
∫(x^2cos(x))/(1+sin(x))^2 dx = ∫(x^2sec(x)) * (1/(1+sin(x))^2) du
= ∫(x^2sec(x))/(u^2) du
7. Now, we can integrate the expression. The integral becomes:
∫(x^2sec(x))/(u^2) du = ∫(x^2/u^2) du
8. Evaluate the integral:
The integral of x^2/u^2 du can be found by using the power rule of integration. We get:
∫(x^2/u^2) du = -x^2/u + 2∫x/u du
= -x^2/u + 2(x/u) + C
9. Finally, substitute back the value of u:
∫(x^2cos(x))/(1+sin(x))^2 dx = ∫(x^2sec(x))/(u^2) du = (-x^2/u + 2(x/u)) + C
Substituting the limits of integration, we get:
= [(-x^2/u + 2(x/u))] from 1 to (1+√2)/√2
Now, plug in the upper limit first and subtract the result when the lower limit is plugged in:
[(-(1+√2)/√2)^2/((1+√2)/√2) + 2((1+√2)/√2)/((1+√2)/√2)] - [(-1^2/1 + 2(1)/1)]
Simplifying the expression will give you the final answer.