Calculus of Trigonometric Functions

Lesson

Integration is typically about finding lengths, areas or volumes by summing a great many small pieces of the whole. Remarkably, where continuous and smooth functions are concerned, it turns out to be the inverse operation of differentiation.

Therefore, integration often begins with the search for an antiderivative.

We know, for example, that $\frac{\mathrm{d}}{\mathrm{d}x}\sin x=\cos x$`d``d``x``s``i``n``x`=`c``o``s``x` and so, $\sin x$`s``i``n``x` is an antiderivative for $\cos x$`c``o``s``x`. An antiderivative is not unique since the derivative of $\sin x+k$`s``i``n``x`+`k`, for any constant $k$`k`, is also $\cos x$`c``o``s``x`. The general antiderivative in this case, $\sin x+k$`s``i``n``x`+`k`, is often called a *primitive*.

For the trigonometric functions, we look for antiderivatives by tracing back from known derivatives. The notation used involves the integral sign, which was introduced by Leibniz in the $17$17th century.

If $f(x)=\sec^2x$`f`(`x`)=`s``e``c`2`x`, find an antiderivative $F(x)=\int\sec^2x\ dx$`F`(`x`)=∫`s``e``c`2`x` `d``x`.

We know from having previously differentiated the tangent function, that $F(x)=\tan x+C$`F`(`x`)=`t``a``n``x`+`C` for any constant $C$`C`, is an antiderivative for $\sec^2x$`s``e``c`2`x`.

To find a definite integral, we first find an antiderivative. Then, we subtract its value at the lower terminal of integration from its value at the upper terminal.

Evaluate $\int_0^{\frac{\pi}{2}}\cos t\ dt$∫π20`c``o``s``t` `d``t`.

The antiderivative is $\sin t$`s``i``n``t`. So, we calculate $[\sin x]_0^{\frac{\pi}{2}}=\sin\frac{\pi}{2}-\sin0=1$[`s``i``n``x`]π20=`s``i``n`π2−`s``i``n`0=1.

Antiderivatives can be difficult to find and may not exist in a form that can be written in terms of simple functions. Various substitutions and transformations are available that can be helpful.

Find an antiderivative for $f(x)=\sin x\ \cos x$`f`(`x`)=`s``i``n``x` `c``o``s``x`.

We make use of the identity $\sin2x=2\sin x\ \cos x$`s``i``n`2`x`=2`s``i``n``x` `c``o``s``x`. So, we require $F(x)=\int\frac{1}{2}\sin2x\ dx$`F`(`x`)=∫12`s``i``n`2`x` `d``x`. You should check by differentiating that $F(x)=-\frac{1}{4}\cos2x$`F`(`x`)=−14`c``o``s`2`x`.

Evaluate $\int_0^{\frac{\pi}{4}}\tan t\ dt$∫π40`t``a``n``t` `d``t`.

We write $\tan t$`t``a``n``t` as $\frac{\sin t}{\cos t}$`s``i``n``t``c``o``s``t`. Observing that $-\sin t$−`s``i``n``t` is the derivative of $\cos t$`c``o``s``t`, we make the substitution $u=\cos t$`u`=`c``o``s``t`. Therefore, $\frac{\mathrm{d}u}{\mathrm{d}t}=-\sin t$`d``u``d``t`=−`s``i``n``t`. and we can substitute for $\sin t$`s``i``n``t` and $\cos t$`c``o``s``t` to write the antiderivative $-\int\frac{1}{u}\frac{\mathrm{d}u}{\mathrm{d}t}\ dt$−∫1`u``d``u``d``t` `d``t`.

It is established elsewhere that this is the same as $-\int\frac{1}{u}\ du$−∫1`u` `d``u` and we call upon our knowledge of the logarithm function to recognise that this is $-\ln u$−`l``n``u`.

Therefore, the antiderivative is $-\ln\cos t$−`l``n``c``o``s``t` (which should be checked by differentiating) and we calculate

$-\left[\ln\cos t\right]_0^{\frac{\pi}{4}}=-\left[\ln\frac{1}{\sqrt{2}}-\ln1\right]=-\ln\frac{1}{\sqrt{2}}\approx0.3466$−[`l``n``c``o``s``t`]π40=−[`l``n`1√2−`l``n`1]=−`l``n`1√2≈0.3466

Integrate $\cos6x$`c``o``s`6`x`.

You may use $C$`C` as the constant of integration.

Integrate $-5\cos\left(\frac{x}{4}\right)$−5`c``o``s`(`x`4).

You may use $C$`C` as the constant of integration.

Calculate $\int_{-\frac{\pi}{6}}^{\frac{7\pi}{6}}\cos3xdx$∫7π6−π6`c``o``s`3`x``d``x`.

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply integration methods in solving problems