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Q.

∫cosxsinx−π4dx=

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a

12logsinx−π4−x−π42

b

logcosx−π4+x−π42+c

c

12logsinx+π4+x+π42+c

d

12logcosx+π4+x+π42+c

answer is A.

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Detailed Solution

x−π4=t⇒dx=dt=∫cos⁡(x)sin⁡x−π4dx=∫cos⁡t+π4sin⁡tdt=∫cos⁡t12−sin⁡t12sin⁡tdt=∫12cot⁡t−12dt=12log⁡(sin⁡t)−t2+c=12log⁡sin⁡x−π4−x−π42+c
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