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Today's Calculation Of Integral
2011 Today's Calculation Of Integral
701
Today's calculation of Integral 701
Today's calculation of Integral 701
Source: created by kunny
June 3, 2011
calculus
integration
trigonometry
calculus computations
Problem Statement
Evaluate
∫
π
4
π
2
(
1
+
cos
x
)
{
1
−
tan
2
x
2
tan
(
x
+
sin
x
)
tan
(
x
−
sin
x
)
}
tan
(
x
+
sin
x
)
d
x
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{(1+\cos x)\{1-\tan ^ 2 \frac{x}{2}\tan (x+\sin x)\tan (x-\sin x)\}}{\tan (x+\sin x)}\ dx
∫
4
π
2
π
tan
(
x
+
sin
x
)
(
1
+
cos
x
)
{
1
−
tan
2
2
x
tan
(
x
+
sin
x
)
tan
(
x
−
sin
x
)}
d
x
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