Calculus of Trigonometric Functions

Lesson

We discovered that a special relationship exists when differentiating sine and cosine functions.

We saw that $\frac{d}{dx}\sin\left(x\right)=\cos\left(x\right)$`d``d``x``s``i``n`(`x`)=`c``o``s`(`x`) and $\frac{d}{dx}\cos\left(x\right)=-\sin\left(x\right)$`d``d``x``c``o``s`(`x`)=−`s``i``n`(`x`)

Since integration reverses the process of differentiation, we can deduce that:

State a primitive function of $6\sin x-\cos x$6`s``i``n``x`−`c``o``s``x`.

You may use $C$`C` as a constant.

Integrate $9\sin3x$9`s``i``n`3`x`.

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

We know from our experience with differentiation, that the derivative of $\cos3x$`c``o``s`3`x` is $-3\sin3x$−3`s``i``n`3`x`. So we have this factor of $3$3 in our expression.

Ignoring the $9$9 in our question, if we were just looking at integrating $\sin3x$`s``i``n`3`x`, then we need to remove this factor of $3$3 by dividing through by $3$3.

Putting it all together we have:

In general then:

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

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

Evaluate $\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\left(4\cos x+\cos4x\right)dx$∫π6−π6(4`c``o``s``x`+`c``o``s`4`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