© examsiri.com
Question : 75 of 160
Marks:
+1,
-0
Solution:
‌‌ Let ‌I=∫‌‌‌ put ‌x=4‌tan‌θ⇒dx=4 sec2θdθ‌∴I‌=∫‌| 4 sec2θdθ |
| (16tan2θ+9)√16tan2θ+16 |
‌=∫‌| 4 sec2θdθ |
| (16‌tan‌‌2‌θ+9)4 secθ |
‌=∫‌=‌∫‌| cos‌θ‌d‌θ |
| 16sin‌2θ+9cos2θ |
=∫‌| cos‌θ‌d‌θ |
| 7sin‌2θ+9 |
Again put
sin‌θ=t‌⇒cos‌θ‌d‌θ=dt‌∴I=∫‌‌‌=‌⋅‌tan−1(‌)+C‌‌=‌tan−1(‌)+C‌‌=‌tan−1(‌)+C‌‌=‌tan−1(‌)+C‌(∵x=4‌tan‌θ⇒sin‌θ=‌)‌∴‌‌K=‌
© examsiri.com
Go to Question: