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Evaluate the integral \int\left (cos\ 2x\ cos\ 3x-sin\ 2x\ sin\ 3x\ \right ) dx

Option: 1

-\frac{1}{5}cos\ 5x+C


Option: 2

\frac{1}{5}cos\ 5x+C


Option: 3

-\frac{1}{5}sin\ 5x+C


Option: 4

\frac{1}{5}sin\ 5x+C


Answers (1)

best_answer

Given integral,

\int\left (cos\ 2x\ cos\ 3x-sin\ 2x\ sin\ 3x\ \right ) dx

\int\left (cos\ 2x\ cos\ 3x+sin\ 2x\ sin\ 3x\ \right ) dx=\int\left (cos\ 2x+3x \right ) dx

\int\left (cos\ 2x\ cos\ 3x+sin\ 2x\ sin\ 3x\ \right ) dx=\int\left (cos\ 5x \right ) dx

Let’s assume,

u=5x\Rightarrow du=5\ dx

dx=\frac{du}{5}

Therefore,

\int\left (cos\ 2x\ cos\ 3x+sin\ 2x\ sin\ 3x\ \right ) dx=\int \frac{1}{5}cos\ u\ du

\int\left (cos\ 2x\ cos\ 3x+sin\ 2x\ sin\ 3x\ \right ) dx=\frac{1}{5}\cdot sin\ u+C

\int\left (cos\ 2x\ cos\ 3x+sin\ 2x\ sin\ 3x\ \right ) dx=\frac{1}{5}sin\ 5x+C

Posted by

Pankaj

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