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Question

(sinx+cosx)4dx=

A
2(tanx+1)2[13(tanx+1)12]+c
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B
2(tanx+1)2[13(tanx+1)+12]+c
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C
2(tanx+1)2[13(tanx+1)+12]+c
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D
2(tanx+1)2[13(tanx+1)12]+c
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Solution

The correct option is A 2(tanx+1)2[13(tanx+1)12]+c
I=(sinx+cosx)4dx=dx[cosx(tanx+1)]4=dxcos2x(tanx+1)4=sec2xdx(tanx+1)4
Put tanx+1=y12tanxsec2xdx=dysec2xdx=2tanxdy=2(y1)dyI=1y4.2(y1)dy=2(1y31y4)dy=2[12y2+13y3]+c=2y2[13y12]+c=2(tanx+1)2[13(tanx+1)12]+c

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