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Question
The value of limx→−π∫sinx0sin−1tdt(x+π)2 is equal to
The value of limx→−π∫sinx0sin−1tdt(x+π)2 is equal to
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Solution
The correct option is B −∞
Given,
∫sin(x)0sin−1(t)(x+π)2dt
=1(x+π)2⋅∫sin(x)0sin−1(t)dt
=1(x+π)2[tsin−1(t)−∫t√1−t2dt]sin(x)0
=1(x+π)2[tsin−1(t)−(−√1−t2)]sin(x)0
=1(x+π)2[tsin−1(t)+√1−t2]sin(x)0
=1(x+π)2(cos(x)+sin(x)sin−1(sin(x))−1)
limx→−π∫sin(x)0sin−1(t)(x+π)2dt
=limx→−π(1(x+π)2(cos(x)+sin(x)sin−1(sin(x))−1))
=∞⋅(−2)=−∞
Given,
∫sin(x)0sin−1(t)(x+π)2dt
=1(x+π)2⋅∫sin(x)0sin−1(t)dt
=1(x+π)2[tsin−1(t)−∫t√1−t2dt]sin(x)0
=1(x+π)2[tsin−1(t)−(−√1−t2)]sin(x)0
=1(x+π)2[tsin−1(t)+√1−t2]sin(x)0
=1(x+π)2(cos(x)+sin(x)sin−1(sin(x))−1)
limx→−π∫sin(x)0sin−1(t)(x+π)2dt
=limx→−π(1(x+π)2(cos(x)+sin(x)sin−1(sin(x))−1))
=∞⋅(−2)=−∞
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