Answer to Question #104846 in Calculus for SHIVAM KUMAR

Question #104846
Integrate w.r.t to x using substitution
(A) ∫ x^3√x^4+1 dx from 1 to 0
(B) tan^7x sec^2x dx from pi by 4 to 0
(C) sec^2 (cosx) sinx dx
1
Expert's answer
2020-03-10T10:06:21-0400

A.)10x3x4+1dxA.) \int_1^0 x^3 \sqrt{x^4 + 1} \, dx

Let us assume that:

x4+1=u4x3dx=dux^4 + 1 = u \\ \Rightarrow 4x^3 \, dx = du

x3dx=du4\Rightarrow x^3 \, dx =\frac{du}{4}

When:

x=1x = 1

Then:

u=14+1=2u = 1^4 + 1 = 2

Again, when:

x=0x = 0

Then:

u=04+1=1u = 0^4 + 1 = 1

Substituting u=x4+1u = x^4 + 1 into the integration and integrating with respect to dudu , we have:

=1421udu= \frac{1}{4} \int_2^1 \sqrt{u} \, du

=1421u12du= \frac{1}{4} \int_2^1 u^{\frac{1}{2}} \, du

=14[u3232]21= \frac{1}{4} \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right]_2^1 [xndx=xn+1n+1]\hspace{1 cm} \left[ \because \int x^n \, dx = \frac{x^{n+1}}{n+1} \right]

=16[u32]21= \frac{1}{6} \left[ u^{\frac{3}{2}} \right]_2^1

=16[132232]=1226= \frac{1}{6} \left[ 1^{\frac{3}{2}} - 2^{\frac{3}{2}} \right] \\ = \frac{1 - 2\sqrt{2}}{6}

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B.) π40tan7(x)sec2(x)dx\int_{\frac{\pi}{4}}^0 \tan^7 (x) \sec^2 (x) \, dx

Let us assume that:

tan(x)=usec2(x)dx=dutan (x) = u \\ \Rightarrow sec^2 (x) \, dx = du

When:

x=π4x = \frac{\pi}{4}

Then:

u=tan(π4)=1u = \tan (\frac{\pi}{4}) = 1

Again, when:

x=0x = 0

Then:

u=tan(0)=0u = \tan (0) = 0

Substituting u=tan(x)u = \tan (x) into the integration and integrating with respect to dudu , we have:

=10u7du=18[u8]10= \int_1^0 u^7 \, du \\ = \frac{1}{8} \left[ u^8 \right]_1^0 [xndx=xn+1n+1]\hspace{1 cm} \left[ \because \int x^n \, dx = \frac{x^{n+1}}{n+1} \right]

=18= - \frac{1}{8}

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C.) sec2(cos(x))sin(x)dx\int \sec^2 (\cos(x)) \sin (x) \, dx

Let us assume that:

cos(x)=usin(x)dx=du\cos(x) = u \\ \Rightarrow - \sin (x) \, dx = du

Substituting u=cos(x)u = \cos (x) into the integration and integrating with respect to dudu , we have:

=sec2(u)du=tan(u)+C= - \int sec^2 (u) \, du \\ = - tan (u) + C \hspace{1 cm} [sec2(x)dx=tan(x)+C]\left[ \because \int \sec^2 (x) \, dx = \tan (x) + C \right]

Where C is a constant of integration.

Undo substitution and we have:

=tan(cos(x))+C= - \tan (\cos(x)) + C [u=cos(x)]\hspace{1 cm} \left[ \because u = \cos(x) \right]


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