Test Index

NCERT Class XII Mathematics Chapter - - Solutions

Show Para  Hide Para 
..... Miscellaneous Exercise .....
Integrate the functions in Exercises 229 to 252.
© examsnet.com
Question : 239 of 271
Marks: +1, -0
1cos⁡(x+a)cos⁡(x+b)\frac{1}{\cos(x+a)\cos(x+b)}
Solution:  
Let I=∫1cos⁡(x+a)cos⁡(x+b)dxI=\int\frac{1}{\cos(x+a)\cos(x+b)}dx
=1sin⁡(a−b)∫sin⁡[(x+a)−(x+b)]cos⁡(x+a)cos⁡(x+b)dx=\frac{1}{\sin(a-b)}\int\frac{\sin[(x+a)-(x+b)]}{\cos(x+a)\cos(x+b)}dx
=1sin⁡(a−b)∫sin⁡(x+a)cos⁡(x+b)−cos⁡(x+a)sin⁡(x+b)cos⁡(x+a)cos⁡(x+b)dx=\frac{1}{\sin(a-b)}\int\frac{\sin(x+a)\cos(x+b)-\cos(x+a)\sin(x+b)}{\cos(x+a)\cos(x+b)}dx
=1sin⁡(a−b)∫[tan⁡(x+a)−tan⁡(x+b)]dx=\frac{1}{\sin(a-b)}\int[\tan(x+a)-\tan(x+b)]dx
=1sin⁡(a−b)[log⁡sec⁡(x+a)−log⁡sec⁡(x+b)]+C=\frac{1}{\sin(a-b)}[\log\sec(x+a)-\log\sec(x+b)]+C
=1sin⁡(a−b)log⁡∣sec⁡(x+a)sec⁡(x+b)∣+C=\frac{1}{\sin(a-b)}\log\left|\frac{\sec(x+a)}{\sec(x+b)}\right|+C
© examsnet.com
Go to Question:
Advertisements