Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
Advertisements
उत्तर
\[Let, I = \int_0^\frac{\pi}{4} \sin2x \sin3x d x ..................(1)\]
\[ \Rightarrow I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \int_0^\frac{\pi}{4} 2\cos2x\frac{\cos3x}{3}dx\]
\[ \Rightarrow I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \frac{2}{3} \left[ \cos2x\frac{\sin3x}{3} \right]_0^\frac{\pi}{4} + \frac{4}{9} \int_0^\frac{\pi}{4} \sin2x \sin3x d x\]
\[ \Rightarrow I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \frac{2}{3} \left[ \cos2x\frac{\sin3x}{3} \right]_0^\frac{\pi}{4} + \frac{4}{9}I ..............\left[From (1) \right]\]
\[ \Rightarrow \frac{5}{9}I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \frac{2}{3} \left[ \cos2x\frac{\sin3x}{3} \right]_0^\frac{\pi}{4} \]
\[ \Rightarrow \frac{5}{9}I = \frac{1}{3\sqrt{2}} + 0\]
\[ \Rightarrow \frac{5}{9}I = \frac{1}{3\sqrt{2}}\]
\[ \therefore I = \frac{3}{5\sqrt{2}}\]
APPEARS IN
संबंधित प्रश्न
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Choose the correct alternative:
`Γ(3/2)`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Which notation represents a definite integral?
