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Re-substituting the value of t, we get∫ dx / √ (a2 –x2) = sin–1 (x/a) + C 6. ∫ dx / √ (x2 + a2) = log |x + √ (x2 + a2)| + C Let’s substitute x = a tan t, so that dx = a sec2 t dt. Therefore,∫ dx / √ (x2 + a2) = ∫ a sec2 t dt...
Compute the following trigonometric integral: A) \int \sin (3x)\cos (x)dx\\ B) \int (\sin x)^{5}(\cos x)^{3}dx\\ C) \int (\tan x)^{5}(\sec x)^{5}dxCalculate the given integral. integral of sin^3 x cos^4 x dxEv...
\int \frac{(e^{2t})}{(e^{t +1})} dt Evaluate: 1) Integration of {(3 / t + 5 / t^3) dt} 2) Integration of {(8 / x + 2e^{-5x}) dx} Integrate: \int_0^{\pi} (1+\cos 5t)^2\sin 5t dt Integrate: \int 3t^2 (t^3 + 4)^5 dt Integrate: \in...
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\int _0^\frac{\pi}{2} \frac{1}{(1+\sin\theta)^3}d\theta=\int_0^\infty\frac{\sec\theta}{(x+\sqrt{1+x^2})^3}dx, \sec\theta=\sqrt{1+x^2} \int_0^\infty \frac{\sqrt{x^2+1}}{(x+\sqrt{1+x^2})^3}dx,t=x+\sqrt{1+x^2} \int_1^\infty \frac{\sqrt{x...
In this article, we discuss an observation phenomenon where the total amount of photons in the full passband of the Birefringent filter is a constant number that is considered by removing the spectrum of the light source irrespective of the instrument transmittance. This conclusion is only noticed ...
Weierstrass Substitution is a helpful tool when we integrate. It can convert trigonometric integrals to integrals of rational function. $$t=\tan(x/2)$$ With this substitution, the trigonometric functions are rational functions of t , $$\sin(x)=\frac{2t}{1+t^2} \quad \cos(x)=\frac{1...
Using integration by parts twice, evaluate the following integral: ∫sin(x)sin(4x)dx. Integration by Parts: Integration by parts is one of the approaches to attain the integration of the form I=∫f(x)g(x)dx. Here it can be noted that two functions f(x) and g(x) ...
t2sin(4t) with respect to dt using integration by parts (IBP). Integration by Parts:Integration by parts is an important method of integration that is used when two functions come together in the integrand. To solve such problems, we select the order of these two functions ...