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Question : 77 of 160
Marks:
+1,
-0
Solution:
Let's consider the given expression as
I=‌‌| sin3x‌cos‌x‌d‌x |
| sin4x+cos4x |
‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) Then
I‌‌=‌‌| sin3(−x)‌cos(−x) |
| sin4(‌−x)+cos4(‌−x) |
dx ‌‌=‌‌| cos3x‌sin‌x |
| cos4x+sin4x |
dx Add equation (I) and (II),
2I‌‌=‌‌| sin3x‌cos‌x+cos3x‌sin‌x |
| cos4x+sin4x |
dx ‌‌=‌‌| sin‌x‌cos‌x(sin2x+cos2x) |
| cos4x+sin4x |
dx ‌‌=‌‌| sin‌x‌cos‌x |
| cos4x+sin4x |
dx ‌‌=‌‌| sin‌x‌cos‌x |
| cos4x(tan4x+1) |
dx Further simplify the above expression,
2I=‌‌| tan‌x‌sec2‌x |
| (tan4x+1) |
dx Let
tan2x=t then
2‌tan‌x‌sec2‌x‌d‌x=dt Now
2I‌‌=‌‌‌ ‌‌=‌[tan−1t]0∞ ‌‌=‌[tan−1(∞)−tan−1(0)] ‌‌=‌×‌ Thus,
2I=‌ I=‌
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