∫0π/2 4sin⁡x+3cos⁡xsin⁡x+cos⁡xdx=

I=∫0π/2 4sin⁡x+3cos⁡xsin⁡x+cos⁡xdx=∫0π/2 4sin⁡(π/2−x)+3cos⁡(π/2−x)sin⁡(π/2−x)+cos⁡(π/2−x)dx=∫0π/2 4cos⁡x+3sin⁡xsin⁡x+cos⁡xdx⇒2I=∫0π/2 7sin⁡x+7cos⁡xsin⁡x+cos⁡xdx=7∫0π/2 sin⁡x+cos⁡xsin⁡x+cos⁡xdx=7[x]=7π2. ∴ I=7π4.

∫0π/2 4sin⁡x+3cos⁡xsin⁡x+cos⁡xdx=

I=∫0π/2 4sin⁡x+3cos⁡xsin⁡x+cos⁡xdx=∫0π/2 4sin⁡(π/2−x)+3cos⁡(π/2−x)sin⁡(π/2−x)+cos⁡(π/2−x)dx=∫0π/2 4cos⁡x+3sin⁡xsin⁡x+cos⁡xdx⇒2I=∫0π/2 7sin⁡x+7cos⁡xsin⁡x+cos⁡xdx=7∫0π/2 sin⁡x+cos⁡xsin⁡x+cos⁡xdx=7[x]=7π2. ∴ I=7π4.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\int_{0}^{π / 2} \frac{4 \sin x + 3 \cos x}{\sin x + \cos x} d x =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$5 π / 4$

$3 π / 2$

$7 π / 4$

$\frac{5 π}{6}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} I = \int_{0}^{π / 2} \frac{4 \sin x + 3 \cos x}{\sin x + \cos x} d x \\ = \int_{0}^{π / 2} \frac{4 \sin (π / 2 - x) + 3 \cos (π / 2 - x)}{\sin (π / 2 - x) + \cos (π / 2 - x)} d x \\ = \int_{0}^{π / 2} \frac{4 \cos x + 3 \sin x}{\sin x + \cos x} d x \\ \Rightarrow 2 I = \int_{0}^{π / 2} \frac{7 \sin x + 7 \cos x}{\sin x + \cos x} d x \\ = 7 \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sin x + \cos x} d x = 7 [x] = \frac{7 π}{2} . \\ ∴ I = \frac{7 π}{4} . \end{array}$