Search any question & find its solution
Question:
Answered & Verified by Expert
If linear function \( f(x) \) and \( g(x) \) satisfy
\( \int[(3 x-1) \cos x+(1-2 x) \sin x] d x=f(x) \cos x+g(x) \sin x+C \), then
Options:
\( \int[(3 x-1) \cos x+(1-2 x) \sin x] d x=f(x) \cos x+g(x) \sin x+C \), then
Solution:
1080 Upvotes
Verified Answer
The correct answer is:
\( g(x)=3(x-1) \)
Given that
$I=\int[(3 x-1) \cos x+(1-2 x) \sin x] d x$
$=f(x) \cos x+g(x) \sin x+c$
$\Rightarrow I=(3 x-1) \sin x-\int \sin x \cdot 3 d x+(1-2 x)(-\cos x)+\int \cos x(-2) d x$
$=(3 x-1) \sin x+3 \cos x-\cos x+2 x \cos x-2 \sin x+c$
$=(3 x-1-2) \sin x+(2+2 x) \cos x+c$
$=3(x-1) \sin x+2(x+1) \cos x+c$
Therefore, $f(x)=2(x+1)$ and $g(x)=3(x-1)$
$I=\int[(3 x-1) \cos x+(1-2 x) \sin x] d x$
$=f(x) \cos x+g(x) \sin x+c$
$\Rightarrow I=(3 x-1) \sin x-\int \sin x \cdot 3 d x+(1-2 x)(-\cos x)+\int \cos x(-2) d x$
$=(3 x-1) \sin x+3 \cos x-\cos x+2 x \cos x-2 \sin x+c$
$=(3 x-1-2) \sin x+(2+2 x) \cos x+c$
$=3(x-1) \sin x+2(x+1) \cos x+c$
Therefore, $f(x)=2(x+1)$ and $g(x)=3(x-1)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.