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Question : 144 of 150
Marks:
+1,
-0
Solution:
∫−11 dx =
∫−11 dx
+
∫−11 dx
I =
I1+I2 I1 =
∫−11 dx
𝐼1 → odd function 𝑓(−𝑥) = −𝑓(𝑥)
so,
I1 = 0
I2 =
∫−11 dx
=
∫−11 dx
→ even function, as f(−x) = f(x)
so,
I2 =
2∫01 dx
=
for x > 0
So, so,
I2 =
2∫01 dx
I2 =
2[ln(x+1)]0−1 𝐼2 = 2ln 2
𝐼 =
𝐼1+𝐼2 ⇒ 𝐼 = 0 + 2ln 2 = 2ln 2
Now, Now
∫0 ln(sin𝛼, )d𝛼 =
−ln2 So,
−∫0 ln(sin𝛼)d𝛼 = 2ln 2 = 𝐼
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