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Question : 73 of 160
Marks:
+1,
-0
Solution:
Let
I=∫‌| 1 |
| cos‌4‌x‌cos‌2‌x |
‌dx ‌I=2‌∫‌| 1 |
| 2‌cos‌4‌x‌cos‌2‌x |
‌dx‌‌=2‌∫‌‌dx‌‌=2‌∫‌| 1 |
| 4cos32x−2‌cos‌2‌x |
‌dx‌[∴cos‌3‌θ=4cos3θ−2‌cos‌θ] ‌=2‌∫‌| 1 |
| 2‌cos‌2‌x(2cos22x−1) |
‌dx‌=∫‌| 1 |
| cos‌2‌x(2cos22x−1) |
Let
⇒1=(2A+B)cos22x+C‌cos‌2‌x−A⇒A=−1,B=2 and
C=0 ∴‌‌I‌=∫‌| 1 |
| cos‌2‌x(2cos22x−1) |
‌dx‌=∫(‌+‌| 2‌cos‌2‌x |
| 2cos22x−1 |
)‌dx ∴‌‌I‌=∫− sec2x‌dx+∫‌| 2‌cos‌2‌x |
| 1−2sin‌22x |
‌dxI‌=‌‌log| sec2x+tan‌2‌x|+I1 Now,
‌‌I1=∫‌| 2‌cos‌2‌x‌dx |
| 1−2sin‌22x |
Let
sin‌2x=t⇒2‌cos‌2‌x‌dx=dt ∴‌‌I1=∫‌=‌‌∫‌ ⇒‌‌I1=‌⋅‌‌log‌|‌|+C I1=‌‌log‌|‌|+C ‌| sec2x+tan‌2‌x|+C ⇒f(x)=√2sin‌2x and g(x)=| sec2x+tan‌2‌x|
Now,
f(‌)=√2sin‌‌=√2⋅(‌)=‌ and
g(‌)=| sec‌+tan‌| =|2+√3|=2+√3 ∴g(‌)−√2f(‌) ‌=2+√3−√2(‌)‌=2+√3−√3‌=2
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