Multiply and divide by (csc x + cot x); use Substitution. Free math lessons and math homework help from basic math to algebra, geometry and beyond. Students, teachers, parents, and everyone can find solutions to their math problems instantly.

You can also find the answer that wolfram alpha provides by repeating almost exactly the work done in the original post, but instead multiplying and dividing by $\csc x+\cot x$. Even the same $u$ -substitution works except the whole numerator is off by a minus sign, which leads to the extra negative outside the logarithm in the antiderivative.

Since the derivative of −cot(x) - cot (x) is csc2 (x) csc 2 (x), the integral of csc2(x) csc 2 (x) is −cot(x) - cot (x). Free math problem solver answers your algebra, geometry, trigonometry, …

\int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx \sum_{n=0}^{\infty}\frac{3}{2^n} Show More

2017年12月25日 · According to the standard integral results, it is known that $\int \csc^2 x \, dx = -\cot x$ since $\frac{d}{dx} \cot x = - \csc^2 x$. However, supposing that this is not known, is there a better method of evaluating $\int \csc^2 x\, dx$ ?

When you write $\int f(x)\,dx$ and expect to end up with something of the simple form $F(x)+C$, then you are only integrating $f$ over an interval on which $f$ is continuous. $\csc(x)$ has discontinuities at $k\pi$. So what has been done here is valid for say the interval $x\in(0,\pi)$.

2023年1月8日 · Answer: the integral of csc (x) is ln|csc (x) + cot (x)| + C. Firstly, we multiply csc (x) with (csc (x) + cot (x))/ (csc (x) + cot (x)) and use the substitution method where u = csc (x) + cot (x). The rest follows.

Evaluation the indefinite integral of csc (x) with detailed explanations.

2024年11月1日 · In this article, we will explore how to find the integral of the cosecant function, \csc (x) csc(x). The integral of \csc (x) csc(x) is less straightforward than those of \sin (x) sin(x) …

2015年11月2日 · The derivative of #cotx# is #csc^2x#, so the integral of #csc^2x# is #cotx + C#. If you really want the integral of the integral, then use: #int[int (cscx)^2 dx] dx = int [-cotx+C] dx# where #C# is an arbitrary constant # = int (-cosx/sinx + C) dx# where #C# is an arbitrary constant # = -lnabssinx + Cx +D# where #C,D# are arbitrary constants