© examsiri.com
Question : 82 of 150
Marks:
+1,
-0
Solution:
Let
II‌=∫‌| sin‌x+sin‌3x |
| cos‌2‌x |
‌dx‌=∫‌| sin‌x(1+sin‌2x) |
| cos‌2‌x |
‌dx‌=∫‌| sin‌x(1+1−cos2x) |
| 2cos2x−1 |
‌dx‌=∫‌| sin‌x(2−cos2x) |
| 2cos2x−1 |
‌dxPut
cos‌x=t⇒sin‌xdx=−dt ∴‌‌I‌=−∫‌‌dt‌=∫‌‌dt‌=‌‌∫‌‌dt‌=‌‌∫(1−‌)‌dt‌=‌‌∫dt−‌‌∫‌‌=‌t−‌⋅‌⋅‌‌log‌|‌|+c‌=‌‌cos‌x−‌‌log‌|‌| √2‌cos‌x−1 |
| √2‌cos‌x+1 |
|+c∴‌‌A‌=‌,B=‌,f(x)=‌| √2‌cos‌x−1 |
| √2‌cos‌x+1 |
© examsiri.com
Go to Question: