Advertisements
Advertisements
प्रश्न
\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx }\]
बेरीज
Advertisements
उत्तर
\[\int\] sin–1 (3x – 4x3)dx
Let x = sin θ
⇒ dx = cos θ.dθ
& θ = sin–1 x
\[\int\] sin–1 (3x – 4x3)dx =
\[\int\] sin–1 (3 sin θ – 4 sin3 θ) . cos θ dθ
= ∫ sin–1 (sin 3θ) . cos θ dθ
\[= 3\int \theta_I . \cos _{II} \theta d\theta\]
\[ = 3\left[ \theta\int\cos \theta d\theta - \int\left\{ \frac{d}{d\theta}\left( \theta \right) - \int\cos \theta d\theta \right\}d\theta \right]\]
\[ = 3\left[ \theta . \sin \theta - \int1 . \sin \theta d\theta \right]\]
\[ = 3\left[ \theta . \sin \theta + \cos \theta \right] + C\]
\[ = 3\left[ \theta . \sin \theta + \sqrt{1 - \sin^2 \theta} \right] + C\]
\[ = 3\left[ \left( \sin^{- 1} x \right) . x + \sqrt{1 - x^2} \right] + C \left( \because \theta = \sin^{- 1} x \right)\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
\[\int\frac{1}{1 + \cos 2x} dx\]
` ∫ 1/ {1+ cos 3x} ` dx
\[\int\frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} dx\]
` ∫ sin x \sqrt (1-cos 2x) dx `
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]
\[\int\frac{1}{1 + \sqrt{x}} dx\]
\[\int x^3 \cos x^4 dx\]
\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]
\[\int \cot^5 x \text{ dx }\]
\[\int \sin^3 x \cos^6 x \text{ dx }\]
\[\int \cos^7 x \text{ dx } \]
\[\int \sin^7 x \text{ dx }\]
\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 - 1}} dx\]
\[\int\frac{3 x^5}{1 + x^{12}} dx\]
\[\int\frac{2x + 5}{x^2 - x - 2} \text{ dx }\]
\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]
\[\int\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} dx\]
\[\int x^2 \sin^2 x\ dx\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int \log_{10} x\ dx\]
` ∫ x tan ^2 x dx
\[\int x \sin x \cos 2x\ dx\]
\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]
\[\int\sqrt{2x - x^2} \text{ dx}\]
\[\int\left( x + 2 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\frac{1}{x\left[ 6 \left( \log x \right)^2 + 7 \log x + 2 \right]} dx\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{x^4}{\left( x - 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to
\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int \sec^4 x\ dx\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]
\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]
