In mathematics, the integral is used to find a function whose derivative is given. In calculus, the integral is used to describe and determine the area of the region limited by the graph of functions. It is also used to evaluate the area under the curve.

Integral is one of the topics of calculus while derivative is also a calculus topic. both the topics are used to find the complex problems of mathematics and physics.

In this article, we will learn about the integral definition, working, and how to solve advanced integral problems with a lot of examples.

**What is integral?**

In mathematics, the integral is the inverse process of derivatives. The term integral is used to evaluate the antiderivative of the given function. Its main work is to calculate area, volume, and many other terms by using graphs.

Integral is divided into two types; one is indefinite integral and the other is a definite integral. When we have to solve the integral without using any limit, we use an indefinite integral.

While we have to apply limits to the calculated integral function either upper or lower limit then it is the definite integral

There are many different techniques and methods to solve the problems of integration. The advanced integral problems are rather tough because we have to follow large calculations for the solution of the advanced problems.

**How to solve advanced integral or antiderivative problems?**

For the solution of advanced integral problems, we must be familiar with the rules of integration. You can use an antiderivative calculator for the solution of advanced integral problems to get the result according to the rules of integration.

**Example 1: For indefinite integral**

Solve 2x^{3} + sin(x) – 12x^{5}y + cos(x) by using indefinite integral, with respect to x.

**Solution **

**Step 1:** Write the given function with integral notation.

ʃ (2x^{3} + sin(x) – 12x^{5}y + cos(x)) dx

**Step 2:** Now apply the sum and the difference rules on the above function.

ʃ (2x^{3} + sin(x) – 12x^{5}y + cos(x)) dx = ʃ (2x^{3}) dx + ʃ sin(x) dx – ʃ 12x^{5}y dx + ʃ cos(x)) dx

**Step 3:** Now take the constants outside the integral notation.

ʃ (2x^{3} + sin(x) – 12x^{5}y + cos(x)) dx = 2ʃ (x^{3}) dx + ʃ sin(x) dx – 12y ʃ x^{5} dx + ʃ cos(x)) dx

**Step 4:** Apply power rule and solve the above equation.

ʃ (2x^{3} + sin(x) – 12x^{5}y + cos(x)) dx = 2(x^{3+1} / 3 + 1) + ʃ sin(x) dx – 12y (x^{5+1} / 5 + 1) + ʃ cos(x)) dx

ʃ (2x^{3} + sin(x) – 12x^{5}y + cos(x)) dx = 2(x^{4} / 4) + ʃ sin(x) dx – 12y (x^{6} / 6) + ʃ cos(x)) dx

ʃ (2x^{3} + sin(x) – 12x^{5}y + cos(x)) dx = 2/4(x^{4}) + ʃ sin(x) dx – 12y/6(x^{6}) + ʃ cos(x)) dx

ʃ (2x^{3} + sin(x) – 12x^{5}y + cos(x)) dx = x^{4}/2 + ʃ sin(x) dx – 2x^{6}y + ʃ cos(x)) dx

**Step 5: **Now apply integral on the trigonometric functions.

ʃ (2x^{3} + sin(x) – 12x^{5}y + cos(x)) dx = x^{4}/2 + (-cos(x)) – 2x^{6}y + sin(x) + C

ʃ (2x^{3} + sin(x) – 12x^{5}y + cos(x)) dx = x^{4}/2 – cos(x) – 2x^{6}y + sin(x) + C

**Example 2**

Solve 21x^{2} + sin(x) * cos(x) – 24xy^{2} by using indefinite integral, with respect to x.

**Solution **

**Step 1:** Write the given function with integral notation.

ʃ (21x^{2} + sin(x) * cos(x) – 24xy^{2}) dx

**Step 2:** Now apply the sum and the difference rules on the above function.

ʃ (21x^{2} + 2sin(x) * cos(x) – 24xy^{2}) dx = ʃ (21x^{2}) dx + ʃ (sin(x) * cos(x)) dx – ʃ (24xy^{2}) dx

**Step 3:** Now take the constants outside the integral notation.

ʃ (21x^{2} + 2sin(x) * cos(x) – 24xy^{2}) dx = 21ʃ (x^{2}) dx + ʃ (sin(x) * cos(x)) dx – 24y^{2} ʃ (x) dx

**Step 4:** Apply integral separately and find I_{1}, I_{2}, and I_{3}.

ʃ (21x^{2} + 2sin(x) * cos(x) – 24xy^{2}) dx = I_{1} + I_{2} – I_{3 }

**For I**_{1}

I_{1} = 21ʃ (x^{2}) dx

I_{1} = 21(x^{2+1} / 2 + 1) + c_{1}

I_{1} = 21(x^{3}/3) + c_{1}

I_{1} = 21x^{3}/3 + c_{1}

I_{1} = 7x^{3} + c_{1}

**For I**_{2}

I_{2} = ʃ (sin(x) * cos(x)) dx

Put t = cos(x), then dt/dx = -sin(x)

dt = -sin(x) dx

put the values in I_{2}

I_{2} = ʃ -t dt

I_{2} = -ʃ t dt

I_{2} = – t^{1+1}/1 + 1 + c_{2}

I_{2} = – t^{2}/2 + c_{2}

Now put the value of t.

I_{2} = – cos^{2}(x)/2 + c_{2}

**For I**_{3}

I_{3} = 24y^{2} ʃ (x) dx

I_{3} = 24y^{2} (x^{1+1}/1 + 1) + c_{3}

I_{3} = 24y^{2} (x^{2}/2) + c_{3}

I_{3} = 24x^{2}y^{2}/2 + c_{3}

I_{3} = 12x^{2}y^{2} + c_{3}

**Step 5:** Now put the values of I_{1}, I_{2}, and I_{3}.

ʃ (21x^{2} + 2sin(x) * cos(x) – 24xy^{2}) dx = I_{1} + I_{2} – I_{3 }

ʃ (21x^{2} + 2sin(x) * cos(x) – 24xy^{2}) dx = 7x^{3} + c_{1} + – cos^{2}(x)/2 + c_{2} – 12x^{2}y^{2} + c_{3}

ʃ (21x^{2} + 2sin(x) * cos(x) – 24xy^{2}) dx = 7x^{3} + – cos^{2}(x)/2– 12x^{2}y^{2} + C

**Example 3: For definite integral**

Solve 138x^{2} + 2sin(x) – 215x^{4}y by using definite integral, with respect to x having limits 2 to 3.

**Solution **

**Summary **

Now you are witnessed that the advanced integral problems are not difficult. Once you follow the above examples, you can solve any integral problem easily. Definite and indefinite integral problems can also be solved by using an antiderivative calculator with steps with steps to avoid large calculations.