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anemone
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Prove that if $[a,\,b]\subset \left(0,\,\dfrac{\pi}{2}\right)$, $\displaystyle \int_a^b \sin x\,dx>\sqrt{b^2+1}-\sqrt{1^2+1}$.
anemone said:Prove that if $[a,\,b]\subset \left(0,\,\dfrac{\pi}{2}\right)$, $\displaystyle \int_a^b \sin x\,dx>\sqrt{b^2+1}-\sqrt{{\color{red}a}^2+1}$.
The integral of a sine function is -cos(x) + C, where C is a constant. This can be derived using integration by parts or by using the substitution method.
To solve the integral of a cosine function, you can use the substitution method or integration by parts. The result will be sin(x) + C, where C is a constant.
Yes, the integral of a tangent function can be simplified using the substitution method or by using the identity tan(x) = sin(x)/cos(x). The result will be -ln|cos(x)| + C, where C is a constant.
Yes, there is a general formula for the integral of a trigonometric function. It is given by ∫sin(ax + b)dx = -cos(ax + b)/a + C, where C is a constant. This formula can be used for any trigonometric function by using the appropriate substitution.
Integrals involving multiple trigonometric functions can be solved using various techniques such as integration by parts, substitution, or trigonometric identities. It is important to carefully choose the appropriate method based on the given function to simplify the integral and arrive at the correct result.