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Question : 78 of 160
Marks:
+1,
-0
Solution:
∫0π| x‌tan‌x |
| sec‌x+tan‌x |
dx Let
I=∫0π| x‌tan‌x |
| sec‌x+tan‌x |
dx ...(i)
∵f(x)dx=f(a+b−x)dx ⇒I=∫0π| (π−x)‌tan(π−x) |
| sec(π−x)+tan(π−x) |
dx I=∫0π| (π−x)(−tan‌x) |
| −sec‌x−tan‌x |
dx I=∫0π| (π−x)‌tan‌x |
| sec‌x+tan‌x |
dx ...(ii)
Adding Eqs. (i) and (ii), we get
2I=∫0π| x‌tan‌x+(π−x)‌tan‌x |
| sec‌x+tan‌x |
dx 2I=∫0π| π‌tan‌x |
| sec‌x+tan‌x |
dx =‌∫0π(×)dx =‌∫0π| sin‌x(1−sin‌x) |
| 1−sin2x |
dx =‌∫0πdx I=‌∫0π(−)dx ‌∫0π(sec‌x‌tan‌x−tan2x)dx ‌∫0π{sec‌x‌tan‌x−(sec2x−1)}dx I=‌∫0π(sec‌x‌tan‌x−sec2x+1)dx =[sec‌x−tan‌x+x]0π ={(sec‌π−tan‌π+π)− {sec‌0°−tan‌0°+0} ={(−1−0+π)−(1−0+0)} =(π−1−1)=(π−2) I=
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