Find the indefinite integral \int (4x^{5}-6\sqrt{x}+2\cos x-4)dx

Khaleesi Herbert 2021-09-18 Answered
Find the indefinite integral \(\displaystyle\int{\left({4}{x}^{{{5}}}-{6}\sqrt{{{x}}}+{2}{\cos{{x}}}-{4}\right)}{\left.{d}{x}\right.}\)

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Expert Answer

grbavit
Answered 2021-09-19 Author has 16633 answers
Step 1
Consider the given indefinite integral:
\(\displaystyle\int{\left({4}{x}^{{{5}}}-{6}\sqrt{{{x}}}+{2}{\cos{{x}}}-{4}\right)}{\left.{d}{x}\right.}\)
Here, the objective is to evaluate the given indefinite integral.
Step 2
Re-write the given indefinite integral as:
\(\displaystyle{I}=\int{\left({4}{x}^{{{5}}}-{6}\sqrt{{{x}}}+{2}{\cos{{x}}}-{4}\right)}{\left.{d}{x}\right.}\)
Use the sum rule of integration \(\displaystyle\int{\left({f{{\left({x}\right)}}}\pm{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}=\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\pm\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle{I}={4}\int{x}^{{{5}}}{\left.{d}{x}\right.}-{6}\int{x}^{{{\frac{{{1}}}{{{2}}}}}}{\left.{d}{x}\right.}+{2}\int{\cos{{x}}}{\left.{d}{x}\right.}-{4}\int{\left.{d}{x}\right.}\)
Use \(\displaystyle\int{x}^{{{n}}}{\left.{d}{x}\right.}={\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}}+{C}\ {\quad\text{and}\quad}\ \int{\cos{{x}}}{\left.{d}{x}\right.}={\sin{{x}}}+{C}\)
\(\displaystyle{I}={\frac{{{4}{x}^{{{6}}}}}{{{6}}}}-{\frac{{{6}{x}^{{{\frac{{{3}}}{{{2}}}}}}}}{{{\frac{{{3}}}{{{2}}}}}}}+{2}{\sin{{x}}}-{4}{x}+{C}\)
\(\displaystyle{I}={\frac{{{2}{x}^{{{6}}}}}{{{3}}}}-{4}{x}^{{{\frac{{{3}}}{{{2}}}}}}+{2}{\sin{{x}}}-{4}{x}+{C}\)
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