Advertisements
Advertisements
प्रश्न
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
बेरीज
उत्तर
\[\int\sin x .\sin 2x .\text{ sin 3x dx }\]
\[ = \frac{1}{2}\int\left( 2 \sin 2x \cdot \sin x \right) \text{ sin 3x dx }\]
\[ = \frac{1}{2}\int\left[ \text{ cos} \left( 2x - x \right) - \text{ cos } \left( 2x + x \right) \right] \text{ sin 3x dx }.............. \left[ \because \text{ 2 sin A sin B = cos (A - B) - cos (A + B)} \right]\]
\[ \Rightarrow = \frac{1}{2}\int\left[ \cos x - \cos 3x \right] \text{ sin 3x dx }\]
\[ = \frac{1}{2}\int\sin 3x \cdot \text{ cos x dx } - \frac{1}{2}\int\sin 3x \cdot \text{ cos 3x dx }\]
\[ = \frac{1}{4}\int \text{ 2 }\sin 3x \cdot \text{ cos x dx} - \frac{1}{4}\int\text{ 2 }\sin 3x \cdot \text{ cos 3x dx } \]
\[ = \frac{1}{4}\int\left[ \sin 4x + \sin 2x \right]dx - \frac{1}{4}\int\text{ sin 6x dx } ............. \left[ \because \text{ 2 sin A cos B = sin (A + B) - sin (A - B)} \right]\]
\[ = \frac{1}{4}\left[ \frac{- \cos 4x}{4} - \frac{\cos 2x}{2} \right] - \frac{1}{4}\left[ - \frac{\cos 6x}{6} \right] + C\]
\[ = - \frac{\cos 4x}{16} - \frac{\cos 2x}{8} + \frac{\cos 6x}{24} + C\]
\[ = \frac{1}{2}\int\left( 2 \sin 2x \cdot \sin x \right) \text{ sin 3x dx }\]
\[ = \frac{1}{2}\int\left[ \text{ cos} \left( 2x - x \right) - \text{ cos } \left( 2x + x \right) \right] \text{ sin 3x dx }.............. \left[ \because \text{ 2 sin A sin B = cos (A - B) - cos (A + B)} \right]\]
\[ \Rightarrow = \frac{1}{2}\int\left[ \cos x - \cos 3x \right] \text{ sin 3x dx }\]
\[ = \frac{1}{2}\int\sin 3x \cdot \text{ cos x dx } - \frac{1}{2}\int\sin 3x \cdot \text{ cos 3x dx }\]
\[ = \frac{1}{4}\int \text{ 2 }\sin 3x \cdot \text{ cos x dx} - \frac{1}{4}\int\text{ 2 }\sin 3x \cdot \text{ cos 3x dx } \]
\[ = \frac{1}{4}\int\left[ \sin 4x + \sin 2x \right]dx - \frac{1}{4}\int\text{ sin 6x dx } ............. \left[ \because \text{ 2 sin A cos B = sin (A + B) - sin (A - B)} \right]\]
\[ = \frac{1}{4}\left[ \frac{- \cos 4x}{4} - \frac{\cos 2x}{2} \right] - \frac{1}{4}\left[ - \frac{\cos 6x}{6} \right] + C\]
\[ = - \frac{\cos 4x}{16} - \frac{\cos 2x}{8} + \frac{\cos 6x}{24} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
` ∫ 1/ {1+ cos 3x} ` dx
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]
\[\int \text{sin}^2 \left( 2x + 5 \right) \text{dx}\]
\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]
\[\int\frac{1}{\sqrt{1 + \cos x}} dx\]
\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]
\[\int\frac{x \sin^{- 1} x^2}{\sqrt{1 - x^4}} dx\]
\[\int\frac{\cos^5 x}{\sin x} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
` = ∫1/{sin^3 x cos^ 2x} dx`
\[\int\frac{x^2 - 1}{x^2 + 4} dx\]
\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]
\[\int\frac{x - 1}{3 x^2 - 4x + 3} dx\]
\[\int\frac{\left( 1 - x^2 \right)}{x \left( 1 - 2x \right)} \text
{dx\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]
\[\int\frac{2x + 5}{\sqrt{x^2 + 2x + 5}} dx\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx }\]
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} dx\]
`int"x"^"n"."log" "x" "dx"`
\[\int\frac{\sin^{- 1} x}{x^2} \text{ dx }\]
\[\int x \sin x \cos 2x\ dx\]
\[\int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
If \[\int\frac{2^{1/x}}{x^2} dx = k 2^{1/x} + C,\] then k is equal to
\[\int\frac{1}{7 + 5 \cos x} dx =\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} \text{ dx }\]
\[\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }\]
\[\int\frac{1}{3 x^2 + 13x - 10} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx}\]
\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx}\]
\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]
\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]
