How do you find #int xe^x - sec(7x)tan(7x)#?

Question

How do you find #int xe^x  -  sec(7x)tan(7x)#?

Emma Aldridge · Accepted Answer

I found: #xe^x-e^x-1/7*1/cos(7x)+c# Let us write it as: #intxe^xdx-intsec(x)tan(x)dx=# we can use Integration by Parts for the first and a transformation for the second: #=xe^x-int1e^xdx-int1/cos(7x)*sin(7x)/cos(7x)dx=# #=xe^x-e^x-intsin(7x)/cos^2(7x)dx=# we can consider that: #d[cos(7x)]=-7sin(7x)dx# and use this into the second integral and write (substituting and dividing by #7#): #=xe^x-e^x+1/7int(d[cos(7x)])/cos^2(7x)=# now we can integrate the second integral using #cos(7x)# as if it was our variable of integration and using the usual rules of integration to get: #=xe^x-e^x-1/7*1/cos(7x)+c#

Aurora Alvarado · Answer

undefined To find the integral of $ \int xe^x - \sec(7x)\tan(7x) $, you can use integration by parts for the first term and the trigonometric substitution for the second term.

1. For $ \int xe^x $, use integration by parts with $ u = x $ and $ dv = e^x \, dx $.
   $$ du = dx \quad \text{and} \quad v = e^x $$
   Apply the integration by parts formula:
   $$ \int u \, dv = uv - \int v \, du $$
   $$ \int xe^x \, dx = xe^x - \int e^x \, dx $$
   $$ = xe^x - e^x + C_1 $$

2. For $ \int \sec(7x)\tan(7x) $, let $ u = \sec(7x) $.
   $$ du = 7\sec(7x)\tan(7x) \, dx $$
   This can be simplified by dividing both sides by $ 7 $:
   $$ \frac{1}{7} du = \sec(7x)\tan(7x) \, dx $$
   So, the integral becomes:
   $$ \frac{1}{7} \int u \, du $$
   $$ = \frac{1}{7} \left( \frac{1}{2} u^2 \right) + C_2 $$
   $$ = \frac{1}{14} \sec^2(7x) + C_2 $$

Putting it all together:
$$ \int xe^x - \sec(7x)\tan(7x) \, dx = xe^x - e^x - \frac{1}{14} \sec^2(7x) + C $$
$$ \text{where } C = C_1 + C_2 $$