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Question : 132 of 150
Marks:
+1,
-0
Solution:
Let
‌‌I=(√tan‌x+√cot‌x)‌dx [| ∵‌ if ‌f(x)=f(2a−x) | ‌ |
| ‌ then ‌f(x)‌dx=2‌f(x)‌dx |
] ∴‌‌I=2‌(√tan‌x+√cot‌x)‌dx=2‌(‌+‌)‌dx=2‌‌| (sin‌x+cos‌x) |
| √sin‌x‌cos‌x |
‌dx=2√2‌‌| sin‌x+cos‌x |
| √2sin‌x‌cos‌x |
‌dx (on multiplying numerator and denominator by
√2 )
=2√2‌‌| sin‌x+cos‌x |
| √1−(sin‌x−cos‌x)2 |
‌dx[∵(sin‌x−cos‌x)2=sin‌2x+cos2x.−2sin‌x‌cos‌x∴‌‌2sin‌x‌cos‌x=1−(sin‌x−cos‌x)2] Put
sin‌x−cos‌x=t⇒cos‌x+sin‌x=‌⇒‌‌dx=‌∴‌‌I=2√2‌‌‌=2√2[sin‌−1t]−10=‌2√2[sin‌−10−sin‌−1(−1)]=‌2√2[0+sin‌−1(1)]‌=2√2(‌)=π√2
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