Integral of trigonometric function

In summary, the integral of a trigonometric function is the inverse operation of differentiation and represents the area under a curve of a trigonometric function. To solve an integral of a trigonometric function, integration techniques such as substitution, integration by parts, or trigonometric identities are used. The applications of integral of trigonometric functions include calculating areas, volumes, and other quantities in real-world problems, as well as in mathematical concepts such as Fourier series. While calculators can be used to solve these integrals, a good understanding of the concepts and techniques involved is important. In special cases, such as when dealing with infinite limits or even powers, special techniques or trigonometric identities may be required to solve the integral.
  • #1
anemone
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Prove that if $[a,\,b]\subset \left(0,\,\dfrac{\pi}{2}\right)$, $\displaystyle \int_a^b \sin x\,dx>\sqrt{b^2+1}-\sqrt{1^2+1}$.
 
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anemone said:
Prove that if $[a,\,b]\subset \left(0,\,\dfrac{\pi}{2}\right)$, $\displaystyle \int_a^b \sin x\,dx>\sqrt{b^2+1}-\sqrt{{\color{red}a}^2+1}$.
[TIKZ][scale=3]
\clip (-0.25,-0.25) rectangle (3.5,3.5) ;
\draw (0,0) circle (3cm) ;
\draw [help lines, ->] (-0.5,0) -- (3.25,0) ;
\draw [help lines, ->] (0,-0.5) -- (0,3.25) ;
\coordinate [label=below left:$0$] (O) at (0,0) ;
\coordinate [label=below left:$1$] (I) at (3,0) ;
\coordinate [label=above right:$A$] (A) at (45:3) ;
\coordinate [label=above:$B$] (B) at (3,2.36) ;
\draw [thick, blue] (I) -- node[ right ]{$x$} (B) ;
\draw [thick, red] (3,0) arc (0:45:3cm) ;
\draw (I) -- (O) -- (A) ;
\draw (O) -- (B) ;
\draw [blue] (0.3,0.1) node{$\theta$} ;
\draw [red] (0.55,0.15) node{$x$} ;
\draw [blue] (0.45,0) arc (0:38:0.45cm) ;
\draw [red](0.7,0) arc (0:45:0.7cm) ;[/TIKZ]

If $0 < x < \frac\pi2$ then $x < \tan x$ and so $\arctan x < x$.

In the above diagram, the red arc of the unit circle and the blue vertical line both have length $x$. So the line $0A$ makes an angle $x$ (radians) with the horizontal, and $\theta = \arctan x < x$. Therefore $\dfrac x{\sqrt{x^2+1}} = \sin\theta < \sin x$.

Now integrate from $a$ to $b$, to get \(\displaystyle \int_a^b\!\! \sin x\,dx > \int_a^b\!\!\frac x{\sqrt{x^2+1}\,}dx = \left[\sqrt{x^2+1}\right]_a^b = \sqrt{b^2+1}-\sqrt{a^2+1}.\)
 

FAQ: Integral of trigonometric function

1. What is the integral of a sine function?

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.

2. How do you solve the integral of a cosine function?

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.

3. Can the integral of a tangent function be simplified?

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.

4. Is there a general formula for the integral of a trigonometric function?

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.

5. How do you solve integrals involving multiple trigonometric functions?

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.

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