∫pi/2 0 root of sinx cosx dx
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Step-by-step explanation:
Let I=∫
0
2
π
sinx
+
cosx
sinx
dx ...(i)
→=∫
0
2
π
sin(
2
π
−x)
+
cos(
2
π
−x)
sin(
2
π
−x)
dx
[∵∫
a
b
f(x)dx=∫
a
b
f(a+b−x)dx]
⇒I=∫
0
2
π
cosx
+
sinx
cosx
dx ...(ii)
On adding Equation (i) and (ii), we get
⇒2I=∫
0
2
π
sinx
+
cosx
sinx
+
cosx
dx
⇒2I=∫
0
2
π
1dx⇒2I=[x]
0
2
π
⇒2I=
2
π
⇒I=
4
π
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