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Q.

∫x sinx sec3x dx =

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a

12[sec2x-tanx]+c

b

12[xsec2x-tanx]+c

c

12[xsec2x+tanx]+c

d

12[sec2x+tanx]+c

answer is B.

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Detailed Solution

∫x sinx sec3x dx​=∫x sinx .1cos3x dx​                                  =∫x tanx sec2x dx​                                              integration by parts, let u=x and v=tanx sec2x                                   ​=x ∫sec xtanx secx dx−∫sec xtanx secx dxdx​                                  =xsec2x2−∫sec2x2dx                                  ​=xsec2x2−tanx2+c
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