Trigonometry in Calculus: Integrals and Accumulated Change

Reverse a derivative, then check it

Differentiating $\sin x$ gives $\cos x$. Reversing that observation gives an antiderivative: $\int\cos x\,dx=\sin x+C$. The constant matters because $\sin x$, $\sin x+2$ and $\sin x-7$ all have the same derivative. An indefinite integral describes a family of antiderivatives on an interval.

Do not read the table as six unrelated instructions. Each row reverses a derivative you already know. In particular, the minus signs in the sine, cotangent and cosecant rows can be checked immediately by differentiating.

Integration respects addition and constant multiples. For example, $\int(3\cos x-2\sin x)\,dx=3\sin x+2\cos x+C$. A single constant is enough for the combined family. By contrast, integration does not generally distribute over a product: the integral of $\sin x\cos x$ is not the product of their separate integrals.

Use radians throughout. Keep the interval in view as well: an antiderivative involving tangent is valid on an interval that does not cross one of its poles. Separate intervals may carry different constants. Your most reliable finishing step is simple: differentiate the proposed answer and recover the original integrand. A plausible-looking expression is not a finished solution until this check works.

Two correct answers may look different after a trigonometric identity has been used. Compare their derivatives or their difference before deciding that one is wrong.

Match the inside with its rate of change

A substitution works when a changing inner expression and its derivative appear together. In $\int\cos(3x)\,dx$, the angle changes three times as fast as $x$. Reversing the chain rule therefore needs a compensating factor:

Differentiating the answer restores the factor $3$ and cancels $1/3$. More generally, for a nonzero constant $a$, $\int\sin(ax+b)\,dx=-\cos(ax+b)/a+C$. If $a=0$, the integrand is constant and must be treated separately.

For a less obvious example, $\int2x\cos(x^2)\,dx$, choose $u=x^2$. Then $du=2x\,dx$, so the integral becomes $\int\cos u\,du=\sin u+C=\sin(x^2)+C$. Write the differential step; it is how you know no factor was lost.

Substitution also explains logarithmic trigonometric integrals. Since $\tan x=\sin x/\cos x$, taking $u=\cos x$ gives $du=-\sin x\,dx$, hence $$\int\tan x\,dx=-\ln|\cos x|+C.$$ Similarly, $\int\cot x\,dx=\ln|\sin x|+C$. The absolute value allows the logarithmic formula on any interval where its argument is nonzero, including intervals where that argument is negative.

Two further useful forms follow by creating a numerator that is the derivative of the denominator: $\int\sec x\,dx=\ln|\sec x+\tan x|+C$ and $\int\operatorname{cosec}x\,dx=\ln|\operatorname{cosec}x-\cot x|+C$. Verify each by differentiation at allowed inputs. The method remains the same: find the inside, locate its derivative and track its coefficient.

After integrating in the new variable, return to the original variable for an indefinite integral. If using substitution in a definite integral instead, either change both bounds consistently or return to the original variable before applying them.

Change the expression, keep the same function

An identity can reveal a familiar integrand that is hidden inside a square or product. The key is to make a useful change, not simply to expand everything. For squared sine and cosine, reverse the double-angle formulas:

The right-hand sides are easier to integrate because they are sums of a constant and a cosine. Thus $$\int\sin^2x\,dx=\frac x2-\frac{\sin(2x)}4+C.$$ The denominator $4$ comes from two factors: the identity contributes $1/2$, and integrating $\cos(2x)$ contributes another $1/2$.

For a product, ask whether a direct substitution or a product-to-sum identity is shorter. Since $2\sin x\cos x=\sin(2x)$, one approach gives $\int\sin x\cos x\,dx=-\cos(2x)/4+C$. Substituting $u=\sin x$ instead gives $\sin^2x/2+C$. These expressions differ only by a constant, so they describe the same antiderivative family.

An odd power suggests saving one factor for a differential. Write $\sin^3x=\sin x(1-\cos^2x)$, then put $u=\cos x$. For $\cos^3x$, save a cosine and use $u=\sin x$. This is more useful than treating a cube as three separate integrals.

Products with different angles can use identities such as $2\sin A\cos B=\sin(A+B)+\sin(A-B)$. Check the sign of the difference angle, then integrate each resulting term with its own chain-rule scale factor. Equivalent integrands stay equivalent; antiderivative answers need only agree up to a constant.

A useful decision is to inspect the powers first. Even squares suggest a double-angle identity; an odd sine or cosine power often offers a factor that can become the differential.

Recognise an inverse angle in an integral

The derivatives of inverse trigonometric functions provide two particularly useful patterns. Let $a>0$. Then

The first form has a positive sum in its denominator and no real poles. The second has a square root and is considered on the open interval where the denominator is positive. Match the whole structure, not just the appearance of a square.

To see the first scale factor, put $u=x/a$, so $dx=a\,du$. The denominator becomes $a^2(1+u^2)$, leaving $(1/a)\int du/(1+u^2)$. Differentiating $(1/a)\tan^{-1}(x/a)$ checks the two factors of $a$.

The root form also has a geometric substitution. Put $x=a\sin\theta$ with $-\pi/2<\theta<\pi/2$. Here $\cos\theta>0$, so $\sqrt{a^2-x^2}=a\cos\theta$, and $dx=a\cos\theta\,d\theta$. The integral reduces to $\int d\theta$. The branch condition is what lets you remove the absolute value from $\sqrt{\cos^2\theta}$.

Changing a plus to a minus can change the answer’s type. For example, $1/(a^2-x^2)$ gives a logarithmic antiderivative, not the inverse-tangent pattern: $$\int\frac{dx}{a^2-x^2}=\frac1{2a}\ln\left|\frac{a+x}{a-x}\right|+C.$$ It is valid on intervals avoiding $x=\pm a$. A quick differentiation and domain check are safer than relying on visual resemblance.

For a definite integral, check that its entire interval lies inside the relevant domain before inserting the bounds. An inverse function may have an endpoint value even when the original root denominator is zero there; that situation needs a separate limiting argument.

Let one factor become simpler

Some products do not contain a convenient inner derivative. In $\int x\cos x\,dx$, choosing $u=x$ does not turn the whole integrand into a single function of $u$. Instead, use integration by parts, which reverses the product rule:

Choose the factor you want to differentiate as $u$, and the part you can integrate as $dv$. For $x\cos x\,dx$, take $u=x$, $du=dx$, $dv=\cos x\,dx$, $v=\sin x$. The remaining integral is simpler:

The plus sign at the end results from integrating sine to negative cosine. Differentiate the answer: the extra $\sin x$ terms cancel, leaving $x\cos x$.

The same choice works for $x\sin x$, but $v=-\cos x$. This gives $-x\cos x+\sin x+C$. With a polynomial of higher degree, repeated parts may help because each differentiation lowers its degree. Keep the changed-angle factor when integrating $\cos(ax)$ or $\sin(ax)$.

An inverse trigonometric function can also be the factor to differentiate. In $\int\sin^{-1}x\,dx$, write the unseen second factor as $1$: take $u=\sin^{-1}x$, $dv=dx$. The remaining integral involves $x/\sqrt{1-x^2}$, which is suited to substitution. Thus methods can cooperate. Choose the next step that simplifies the remaining work, then check the final result by differentiation. State the interval where the inverse derivative or denominator used in the method is valid.

Read a definite integral as accumulated change

An indefinite integral gives a family of functions. A definite integral, with fixed lower and upper bounds, gives a number. If $f$ is continuous on $[a,b]$ and $F$ is an antiderivative there, the Fundamental Theorem of Calculus gives

The constant cancels. Evaluate at the upper bound first, subtract the lower-bound value, and keep the angles in radians.

For a graph, the integral is signed accumulation (चिह्न सहित संचय): regions above the horizontal axis contribute positively, and regions below contribute negatively when the bounds run left to right. On $[0,\pi]$, sine is nonnegative and its integral is $2$, equal to the geometric area. On $[\pi,2\pi]$, its integral is $-2$. Therefore $\int_0^{2\pi}\sin x\,dx=0$, although the total geometric area of the two lobes is $4$.

Use the area explorer to move one bound across a zero. Predict whether the accumulated value should increase or decrease before reading it. A zero net integral can represent cancellation rather than an empty region. Reversing the bounds also changes the sign: $\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$.

Check continuity on the whole interval before applying an endpoint formula. Tangent and secant are undefined where cosine is zero. Thus an interval reaching or crossing $\pi/2$ does not meet the continuity condition for directly applying this theorem to $\sec^2x$. On $[0,\pi/4]$, the integrand is continuous and its tangent antiderivative gives a valid finite result.