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Question : 5 of 160
Marks:
+1,
-0
Solution:
Let
I=‌| 16x‌sin‌x‌cos‌x |
| sin4x+cos4x |
dx =‌| 16(−x)‌sin(−x)‌cos(−x) |
| sin4(‌−x)+cos4(‌−x) |
dx =‌| (8π−16x)‌sin‌x‌cos‌x |
| sin4x+cos4x |
dx On adding Eqs. (i) and (ii), we get
2I=8π‌‌| sin‌x‌cos‌x |
| sin4x+cos4x |
dx ∴ =4π‌‌| 2‌sin‌x‌cos‌x |
| sin4x+cos4x |
dx I=2π‌‌| 2‌tan‌x‌sec2‌x |
| tan4x+1 |
dx =2π‌d{tan−1(tan2x)} ‌‌=2π[tan−1(tan2x)]0π∕2=2π[‌−0] ∴‌‌I=π2
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