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Question : 76 of 160
Marks:
+1,
-0
Solution:
‌I=‌‌dx‌I=‌| cos‌θ‌d‌θ |
| (sin‌θ+cos‌θ)cos2θ |
Put
x=sin‌θ⇒dx=cos‌θ‌d‌θ=‌| dθ |
| sin‌θ‌cos‌θ+cos2θ |
Dividing
NT and
Dr by
cos2θ, we get
=‌dθPut
tan‌θ+1=t, sec2θdθ=dt‌=∫‌+12‌=[ln‌t]1+‌2‌=ln‌2−ln(1+‌)‌=ln‌2−ln(‌)=ln(‌)‌=ln(‌)=ln(3−√3)
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